Chapter 4: Motion in Two Dimensions

Transcription

1 Answers and Solutons. (a) Tme s a scalar quantt. (b) Dsplacement s a vector quantt. (c) Veloct s a vector quantt. (d) Speed s a scalar quantt.. An nspecton of the lengths, or magntudes, or the vectors n the fgure leads to the rankng, < C < A < D. 3. Pcture the Problem: Ths s a follow up queston to Guded Eample 4.. At a local sk resort ou take a rope tow to the top of the bunn slope. Your dsplacement vector for ths trp has a length of 90 m and ponts at an angle of 6 above the horzontal. Strateg: Decreasng the angle of the char lft would make the dsplacement vector more horzontal. Use the cosne functon to determne the new horzontal dstance d when the angle s decreased. Soluton:. (a) Decreasng the angle of the char lft makes the dsplacement vector more horzontal, ncreasng d and decreasng d. We conclude that f the angle of the char lft s decreased, the horzontal dstance d wll ncrease.. (b) Use the cosne functon to fnd d : d d ( ) ( ) = cos = 90 m cos 5 = 84 m 80 m Insght: As epected, d ncreased from 7 m (rounded to 70 m n the eample because d = 90 m has onl two sgnfcant fgures) when = 6 to 84 m when = (a) An nspecton of the two vectors reveals that vector ponts more horzontall than vector A, and thus has the greater component. (b) An nspecton of the two vectors reveals that vector A ponts more vertcall than vector, and thus has the greater component. 5. Pcture the Problem: A poston vector has a gven magntude and angle relatve to the as. Strateg: Use the cosne and sne functons to determne the and components, respectvel. Soluton:. (a) Use the cosne functon to fnd r : r r ( ) ( ) = cos = 75.0 m cos 35 = 6.4 m. Use the sne functon to fnd r : r r ( ) ( ) = sn = 75.0 m sn 35 = 43.0 m 3. (b) Use the cosne functon to fnd r : r r ( ) ( ) = cos = 75.0 m cos 65 = 3.7 m 4. Use the sne functon to fnd r : r r ( ) ( ) = sn = 75.0 m sn 65 = 68.0 m Insght: In part (a) the 35 angle means the vector ponts more horzontall than vertcall, and we see that the component s greater than the component. In part (b) the 65 angle means the vector ponts more vertcall than horzontall, so that the component s greater than the component. 6. Pcture the Problem: A press bo s 9.75 m above second base and an unknown horzontal dstance awa. Strateg: Use the tangent functon to determne the horzontal dstance. Soluton: Use the tangent functon to fnd : 9.75 m = = = 36.4 m tan tan5.0 Insght: Alternatvel, ou could use the sne functon to fnd the total dstance (the hpotenuse of the trangle) s = 37.7 m cos5.0 = 36.4 m. d = ( 9.75 m) sn5.0 = 37.7 m, and then use the cosne functon to fnd ( ) m Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4

2 Pearson Phscs b James S. Walker 7. Pcture the Problem: Ths s a follow up queston to Guded Eample 4.3. A pece of plwood s cut to make a skateboard ramp that covers a horzontal dstance of.33 m and rses a vertcal dstance of m. Strateg: Use the relatonshp among the angle, the horzontal dstance, and the vertcal dstance to answer the queston about how the angle changes when the horzontal dstance s doubled and the vertcal rse remans the same. Soluton:. (a) Use a sketch to fnd that the angle wll decrease f the horzontal dstance s doubled but the vertcal rse remans the same. æd ö - -æ0.380 mö. (b) Use the tangent functon to fnd : = tan tan = = 8.3 ç èd çè.66 m Insght: Note that the angle was nearl, but not qute, cut n half from the 5.9 found n Guded Eample 4.3. If t had been cut eactl n half we would have found = It was not cut eactl n half because the tangent functon s a nonlnear functon. 8. Pcture the Problem: The Lewston grade road gans 6 ft n alttude for ever 00 ft t spans n the horzontal drecton. Strateg: Use the tangent functon to fnd the angle. Soluton: Appl the tangent functon: tan = - 6 ft = tan æ ç ö = 3 çè00 ft 00 ft 6 ft Insght: There s onl one sgnfcant fgure n the gven nformaton. If we knew the road was a 6.00 percent uphll grade, we could fnd the angle s Pcture the Problem: You slde a bo up the ramp that s depcted at rght. Strateg: Use the nverse sne functon to fnd the angle usng the pertnent sdes of the trangle. Soluton:. Wrte the defnton of the sne functon:. Use the nverse sne functon to fnd the angle: opposte sn = hpotenuse = r. m = sn - ç æ ö = 7 çè3.7 m 3.7 m Insght: Fndng a rght trangle n an phscs problem allows ou to use the arsenal of trgonometrc tools to fnd varous other quanttes of nterest. Learn to fnd them!. m 0. Vectors are dstngushed from scalars because the have a specfc drecton assocated wth them, whereas scalars do not.. The magntude of a vector refers to ts length.. The magntude of a vector can be found b addng ts components n quadrature. That s, the sum of the squares of the components s the square of the vector s magntude. In equaton form, r = r + r. 3. To resolve a vector s to fnd the two vectors that represent the components of the orgnal vector. The sum of the component vectors s dentcal to the orgnal vector. The vector components are the lengths of the orgnal vector along specfed drectons. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4

3 Pearson Phscs b James S. Walker 4. Pcture the Problem: Each component of a vector s doubled n magntude. Strateg: Note the relatonshp between the components of a vector and ts magntude and drecton to answer the conceptual queston. Soluton:. (a) Doublng each of the components of a vector wll ncrease ts magntude b a multplcatve factor of. You can pcture ths n our head or confrm t mathematcall wth a calculaton lke the followng: = + ( ) ( ) ( ) A A A A + A = 4 A + A = A + A = A. (b) Doublng each component of a vector wll not change ts drecton at all; the drecton angle wll sta the same. You can pcture ths n our head or confrm t mathematcall wth a calculaton lke the followng: = tan ( A A) tan ( ) tan ( ) A A = A A = Insght: You can change a vector s drecton onl b changng the relatve magntudes of ts components. In ths eercse each component was changed b the same multplcatve factor, so the relatve magntudes were unchanged. 5. Pcture the Problem: An arplane drops a vertcal dstance of 4 m and moves forward a horzontal dstance of 30 m. Strateg: Use the Pthagorean theorem to determne the total dstance. Soluton: Use the Pthagorean theorem to fnd d: d = d + d ( ) ( ) = 30 m + 4 m = 3 m Insght: Note that durng ths gentle descent (the angle ( ) nearl dentcal to the horzontal dstance. d = 30 m = tan 4 m 30 m = 4.3 ) the total dstance traveled s d = 4 m 6. Pcture the Problem: A d =.4-km-long nclned road rses d = 60 m n elevaton. Strateg: Use the sne functon to determne the angle. Soluton: Use the sne functon to fnd, notng that.4 km 000 m/km = 400 m: sn = d d 60 m 400 m = sn = 3.8 Insght: For small angles lke ths one, the horzontal dstance s nearl equal to the total dstance. If the total dstance were eactl 400 m n ths problem, the horzontal dstance would be 395 m. d d 7. Pcture the Problem: The length and drecton of a dsplacement vector are gven. Strateg: Resolve d nto ts and components b usng the sne and cosne functons. In ths problem the drecton s east, the drecton s north, and n the dagram represents the angle north of east that s gven n the problem. Soluton:. Fnd d : ( ) d = 760 m cos35 = 60 m d d d. Fnd d : ( ) d = 760 m sn 35 = 440 m Insght: An vector can be resolved nto two components. The ablt to convert a vector to and from ts components s an essental skll for solvng man phscs problems. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 3

4 Pearson Phscs b James S. Walker 8. Pcture the Problem: The horzontal and vertcal components of a dsplacement vector are gven. Strateg: Use the and components of d to fnd the drecton angle. In ths problem the drecton s horzontal and the drecton s vertcall upward. Soluton: Use the tangent functon to fnd : æd ö - -æ4.8 mö = tan = tan = ç è d èç m d d d Insght: An vector can be resolved nto two components. The ablt to convert a vector to and from ts components s an essental skll for solvng man phscs problems. 9. Pcture the Problem: Consder the two vectors A and depcted n the fgure. Strateg: Remember the rules of addng and subtractng vectors. Vectors are alwas added head-to-tal. To subtract vectors, reverse the drecton of the negatve vector and add t head-to-tal to the postve vector. Soluton:. (a) To subtract A we must reverse the drecton of and move t so that ts tal s on the head of A. The resultant vector starts at the tal of A and ends at the head of so t ponts down and to the rght lke vector D.. (b) To subtract A D we must reverse D (so that t ponts up and to the left) and add t head-to-tal to A. The resultant vector ponts up and to the rght, but s farl short because the leftward component of D brngs the resultant closer to ts startng pont. However, to subtract A E we must reverse E (so that t ponts down and to the rght) and add t head-to-tal to A. The resultant vector ponts down and to the rght, and s farl long because the rghtward component of E brngs the resultant farther from ts startng pont. We conclude that the magntude of A D s less than the magntude of A E. Insght: In the fgure the vector F ponts n a smlar drecton to A. 0. In order to correctl add two vectors ou must place the tal of one vector on the head of the other. The resultant vector s the same no matter n whch order the two vectors are added.. In order to correctl add two vectors usng ther components, ou must add the two components together and add the two components together. These sums gve ou the and components, respectvel, of the resultant vector.. Pcture the Problem: Consder the two vectors A and depcted n the fgure. Strateg: Remember the rules of addng and subtractng vectors. Vectors are alwas added head-to-tal. To subtract vectors, reverse the drecton of the negatve vector and add t head-to-tal to the postve vector. Soluton: To add A+ D we must add D head-to-tal to A. The resultant vector ponts down and to the rght, and s farl long because the rghtward component of D brngs the resultant farther from ts startng pont. To add A+ E we must add E head-to-tal to A. The resultant vector ponts up and to the rght, but s farl short because the leftward component of E brngs the resultant closer to ts startng pont. We conclude that the magntude of A+ D s greater than the magntude of A+ E. Insght: In the fgure the vector F ponts n a smlar drecton to A. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 4

5 Pearson Phscs b James S. Walker 3. Pcture the Problem: Consder the two vectors A and depcted n the fgure. Strateg: Remember the rules of addng and subtractng vectors. Vectors are alwas added head-to-tal. To subtract vectors, reverse the drecton of the negatve vector and add t head-to-tal to the postve vector. Soluton: To subtract C we must reverse C (so that t ponts down and to the left) and add t head-to-tal to. The resultant vector ponts to the left, but s farl short because the downward component of C brngs the resultant closer to ts startng pont. However, to subtract F we must reverse F (so that t ponts up and to the rght) and add t head-to-tal to. The resultant vector ponts up and to the rght, and s farl long because the upward component of F brngs the resultant farther from ts startng pont. We conclude that the magntude of C s less than the magntude of F. Insght: In the fgure the vector F ponts n a smlar drecton to A. 4. Pcture the Problem: Vectors are sketched to show that two vectors of unequal magntude 0 cannot add to zero, but that three vectors of unequal magntude can. 7 Strateg: Use a vector dagram to show how vectors of dfferent magntudes can be added head-to-tal. 0 Soluton: The two dagrams at the rght llustrate the addton of vectors of dfferent 3 7 magntude. In the top dagram there s a vector of magntude 0 unts that ponts to the rght, and a vector of magntude 7 unts that ponts toward the left. Even f we add the vectors n opposte drectons, a 7-vector can never cancel out a 0-vector, and the resultant wll alwas have a nonzero magntude. The bottom dagram shows a vector of magntude 0 unts that ponts to the rght, and vectors of magntude 7 and 3 unts, respectvel, that pont toward the left. When the are arranged n ths fashon the three vectors sum to zero. Insght: If the vectors all had the same length, two of them would sum to zero f ou added them n opposte drectons, and three of them would sum to zero f ou formed them nto the sdes of an equlateral trangle and the all ponted clockwse or all ponted counterclockwse. 5. Pcture the Problem: The two vectors A (length 7 m) and (length 35 m) are drawn at rght. Strateg: Add vectors A and usng the vector dagram method. Soluton: A sketch of the vectors and ther sum s shown at rght. When A and are added head-to-tal, the resultant C= A+ ponts toward the rght and down. It s somewhat longer than so we estmate ts length s 45 m and t ponts about 0 below the as. Insght: Usng a ruler and protractor to add the vectors graphcall s not as accurate as the component method, but t helps ou to better pcture the stuaton. It s the preferred method when ou frst learn how to add vectors. A 3 C 55 Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 5

6 Pearson Phscs b James S. Walker 6. Pcture the Problem: The two vectors A (length 7 m) and (length 35 m) are drawn at rght. Strateg: Add vectors A and usng the vector component method. A = ( ) = A = ( ) = ( ) ( ) ( ) ( ) Soluton:. Fnd the components of A and : 7 m cos3.9 m 7 m sn m = 35 m cos 55 = 0. m = 35 m sn 55 = 8.7 m A 3 C 55. Add the components: C = A + = m = 43.0 m 3. Add the components: C = A + = m = 4.4 m 4. Fnd the magntude of C : C C C ( ) ( ) = + = 43.0 m m = 45 m 5. Fnd the drecton of C : æc ö - -æ4.4 m ö C = tan tan = = -9 ç C çè43.0 m or 9 below the as. è Insght: Resolvng vectors nto components takes a lttle bt of etra effort, but ou can get much more accurate answers usng ths approach than b usng a ruler and protractor to add the vectors graphcall. 7. Pcture the Problem: The vectors nvolved n the problem are depcted at the rght. Strateg: Use the vector component method of addton and subtracton to determne the components of each combnaton of A and. Once the components are known, the length and drecton of each combnaton can be determned farl easl. 33 A+ A+ A A O A A A Soluton:. (a) Determne the components of A+. Fnd the magntude of A+ 3. Determne the drecton of A+, measured counterclockwse from the postve as. 4. (b) Determne the components of A 5. Fnd the magntude of A 6. Determne the drecton of A, measured counterclockwse from the postve as. 7. (c) Determne the components of A : A+ = ( ) ˆj+ ( ) ˆ= ( ˆ+ ˆj ) m 33 m 33 m + = m + 33 m = 35 m : A ( ) ( ) - æ mö A = tan = 0 cw from - as or 60 + ç çè33 m A = m ˆj 33 m ˆ= 33ˆ+ ˆj m : ( ) ( ) ( ) : A ( ) ( ) = 33 m + m = 35 m - æ mö A = tan = 0 ccw from + as - ç çè33 m A = 33 m ˆ m ˆj = 33ˆ ˆj m : ( ) ( ) ( ) Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 6

7 Pearson Phscs b James S. Walker 8. Fnd the magntude of A 9. Determne the drecton of A, measured counterclockwse from the postve as. : A ( ) ( ) = 33 m + m = 35 m æ mö = tan ç = 0 ccw from - as çè33 m = 00 ccw from + as - - A Insght: Ths problem s smplfed b the fact that A and have onl one component each, but a smlar approach wll work even wth more complcated vectors. Notce that ou must have a pcture of the vectors n our head (or on paper) n order to correctl nterpret the drectons n steps 3, 6, and Pcture the Problem: Ths s a follow up queston to Guded Eample 4.8. You are rdng n a boat whose speed relatve to the water s 6. m/s. The boat ponts at an angle of 5 upstream on a rver flowng at 4.5 m/s. Strateg: To fnd the veloct of the boat relatve to the ground we use v = v + v wth b referrng to the boat, w referrng to the water, and bg bw wg, g referrng to the ground. The dagram llustrates how the vectors add. as needed to correctl nterpret the soluton. Soluton:. (a) Fnd the and components of the veloct of the water relatve to the ground:. Fnd the and components of the veloct of the boat relatve to the water: 3. Add the components of v bw and v wg to fnd the components of v : bg v v v v v v wg, wg, bw, bw, bg, bg, = 0 = 4.5 m/s = ( 6. m/s) cos 5 = 5.5 m/s = ( 6. m/s) sn 5 =.6 m/ s = m/s = 5.5 m/s = m/s =.9 m/s 4. Fnd the magntude of bg bg = bg, + bg, = 5.5 m/s +.9 m/s = 5.8 m/s 5. Fnd the angle of v æv ö - bw,.9 m/s bg from ts components: - æ- ö = tan tan ç = 9 vbw, 5.5 m/s = - çè çè or 9 below the + as. Insght: Even though the boat s ponted upstream, the current s so strong that t s beng carred downstream. v from ts components: v v v ( ) ( ) 9. Pcture the Problem: The vectors nvolved n ths problem are depcted at rght. Strateg: Let v = pg veloct of the plane relatve to the ground, v = pa veloct of the plane relatve to the ar, and v = ag veloct of the ar relatve to the ground. The drawng at rght depcts the vectors added accordng to the equaton, v pg = v pa + v ag. Determne the angle of the trangle from the nverse sne functon. Soluton:. Use the nverse sne functon to fnd : æv - ag ö -æ 65 km/h ö = sn sn = ç = east of north ç v çè pa 340 km/h è. The angle east of north s the same as 90 = 79 north of east. Insght: If the plane s speed were to be reduced to 40 km/h, the requred angle would become 6 east of north. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 7

8 Pearson Phscs b James S. Walker 30. The captan of the boat n Guded Eample 4.8 should decrease ts speed untl the component of the boat s veloct relatve to the water v bw s equal and opposte to the.4 m/s veloct of the water v. wg In that case there wll be no component to the veloct of the boat relatve to the ground v, bg and the boat wll move straght across the rver, parallel to the as. 3. The addton of veloctes s used to fnd relatve moton, so that f the veloct of object relatve to object s v, and the veloct of object relatve to object 3 s v 3, then the veloct of object relatve to object 3 s v = v + v The wnd speed relatve to our bod s greatest when ou run aganst the wnd and least when ou run wth the wnd. For nstance, f ou run at 4 m/s (relatve to the ground) on a da when the wnd s 7 m/s relatve to the ground, the wnd speed relatve to our bod s m/s when ou run nto the wnd and 3 m/s when ou run wth the wnd. 33. In Guded Eample 4.8 the boat s ponted 5 upstream, but onl moves n a drecton that s upstream, because the rver current moves the boat n a drecton (downward n the fgure) that decreases the angle of ts path wth respect to the ground. If there were no current the boat would move n a drecton 5 relatve to the as. 34. Pcture the Problem: A person walks east on a ferr as the ferr moves north. The vectors nvolved n ths problem are depcted at rght. Strateg: Let v pf = the passenger s veloct relatve to the ferr, v = fd the ferr s veloct relatve to the dock, and v = pd the person s veloct relatve to the dock. the addton of veloctes we can see that v pd = v pf + v fd. v fd v pf v pd N E Soluton:. Add the veloct vectors: v = v + v pd pf fd (. m/s) ˆ ( 6. m/s) = + ˆj. Fnd the magntude of pd : v pd =. m/s + 6. m/s = 6.3 m/s v ( ) ( ) Insght: If the person were to walk even faster wth respect to the ferr, then easterl drecton. v pd would be longer and pont more n the 35. Pcture the Problem: A person walks north on a cruse shp as the shp moves east. The vectors nvolved n ths problem are depcted at rght. Strateg: Let v = ps the passenger s veloct relatve to the shp, v = sw the shp s veloct relatve to the water, and v = pw the person s veloct relatve to the water. the addton of veloctes we can see that v pw = v ps + v sw. Soluton:. Add the veloct vectors: v pw = v ps + v sw = ( 3.8 m/s) ˆj+ ( m/s) ˆ æv ö - pw, -æ3.8 m/sö. Use the tangent functon to fnd : = tan tan ç = ç = 8 ç èv çè pw, m/s v pw Insght: If the person were to run even faster wth respect to the ferr, then northerl drecton, so that would ncrease. v p s N E v sw north of east. v would be longer and pont more n the pw Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 8

9 Pearson Phscs b James S. Walker 36. Pcture the Problem: Ths s a follow up queston to Actve Eample 4.9. A person skateboardng wth a constant speed of.30 m/s releases a ball from a heght of.5 m above the ground. Strateg: The tme requred for the ball to reach the ground s eactl the same as the fall tme f t were dropped from rest, because the horzontal and vertcal motons are ndependent. Soluton:. Wrte the vertcal poston equaton for the ball, notng v, = 0: = + 0 gt f. Solve the epresson for t: = gt f ( ) = t f g ( ) ( ) f 0.5 m = t = = s g 9.8 m s Insght: If the person were to double hs horzontal speed to.60 m/s, the fall tme for the ball would reman eactl the same because from the standpont of the vertcal moton, the ball stll falls from rest a dstance of.5 m. 37. Pcture the Problem: An arrow falls below the target center as t fles from the bow to the target. Strateg: Treat the vertcal and horzontal motons separatel. Fnd the tme requred for the arrow to drop straght down 5 cm from rest. v 0.5 m Soluton: Fnd the tme to drop 5 cm: ( ) = + 0 gt f = t f g t ( ) ( ) m f = = = g 9.8 m/s 0.33 s Insght: Note that the 5-m dstance to the target does not factor nto the tme requred for the arrow to fall 0.5 m. However, we could use the dstance to fnd the ntal speed: v =Δ t = ( 5 m) ( 0.33 s) = 45 m/s. 38. Pcture the Problem: Ths s a follow up queston for Quck Eample 4.0. You throw a baseball from the roof of a house to a frend wth an ntal veloct of.0 m/s n the horzontal drecton. Strateg: Fnd the speed of the ball b separatel determnng the horzontal and vertcal components of the ball s veloct. Then add the components usng the Pthagorean theorem to fnd the speed of the ball at t =.00 s. Soluton:. The horzontal veloct remans the same throughout the flght because there s no acceleraton along that drecton:. The vertcal veloct s a result of the acceleraton of gravt: v =.0 m/s v = v gt,f, ( )( ) = m s.00 s = 9.8 m/s Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 9

10 Pearson Phscs b James S. Walker 3. The speed s = + v v v ( ) ( ) v v v : = + =.0 m/s m/s = 5.5 m/s Insght: The vertcal speed wll contnue to ncrease at 9.8 m/s but the horzontal speed wll reman constant (n the absence of eternal forces such as ar frcton) throughout the flght. 39. Pcture the Problem: Ths s a follow up queston for Quck Eample 4.0. You throw a baseball from the roof of a house to a frend wth an ntal veloct of.0 m/s n the horzontal drecton. Strateg: Fnd the drecton of the ball b separatel determnng the horzontal and vertcal components of the ball s veloct. Then use the nverse tangent functon to fnd the drecton of the ball s travel at t =.00 s Soluton:. The horzontal veloct remans the same throughout the flght:. The vertcal veloct s a result of the acceleraton of gravt: 3. The drecton s = ( v v) tan : v =.0 m/s v = v gt,f, ( )( ) = m s.00 s = 9.8 m/s v 9.8 m/s = tan = tan = 39.3 v.0 m/s or 39.3 below the horzontal. Insght: The vertcal speed wll contnue to ncrease at 9.8 m/s but the horzontal speed wll reman constant (n the absence of eternal forces such as ar frcton) throughout the flght. Ths means that wll contnue to ncrease. 40. Pcture the Problem: Ths s a follow up queston for Guded Eample 4.. A golfer sends the ball over a 3.00 m-hgh tree that s 4.0 m awa. The ball passes drectl over the tree at t =.76 s. Strateg: Use the gven tme at whch the ball s drectl above the tree to fnd the horzontal and vertcal components of the ball s veloct at that nstant. Add the components usng the Pthagorean theorem to fnd the speed of the ball. The ntal veloct was determned to be 3.5 m/s n Guded Eample 4.. Soluton:. The horzontal veloct remans the same throughout the flght:. The vertcal veloct s a result of the acceleraton of gravt: 3. The speed s v v v : ( ) v = v cos = 3.5 m/s cos54.0 = m/s v = v sn gt,f ( ) ( )( ) = 3.5 m/s sn m s.76 s = m/s = + = m/s m/s = 0. m/s = + v v v ( ) ( ) Insght: You can use the veloct components to fnd the angle of the veloct at the nstant that the ball clears the tree æv ö - -æ m/sö s = tan tan = = ç v çè m/s or 38.6 below the horzontal. It appears greater than that angle n è the fgure because the horzontal and vertcal scales are dfferent. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 0

11 Pearson Phscs b James S. Walker 4. Pcture the Problem: A soccer ball s kcked from level ground and travels along a parabolc trajector. Strateg: Use the poston-tme equatons for projectles launched at an angle n order to fnd the and postons of the soccer ball at a gven tme. Soluton:. The horzontal veloct remans the same throughout the flght: f ( cos) ( ) ( ) f = + v t = 0 + m/s cos s = 5. m ( ). The vertcal poston f = + vsn t gt s affected b the acceleraton of gravt: = 0 + ( m/s) sn 3 ( 0.50 s) ( 9.8 m s )( 0.50 s ).0 m Insght: The ball wll arrve at the peak of ts flght at t = v sn g = s, so the ball has not et reached the peak. It s peak heght wll be ma =.06 m when ts downfeld poston s = 6.60 m. 4. Pcture the Problem: A soccer ball s kcked from level ground and travels along a parabolc trajector. Strateg: Use the veloct-tme equatons for projectles launched at an angle n order to fnd the and components of the soccer ball s veloct at a gven tme. Soluton:. The horzontal veloct remans the same v v cos,f = throughout the flght: ( ). The vertcal veloct s affected b the acceleraton of gravt: = m/s cos3 = 0 m/s v = v sn gt,f ( ) ( )( ) = m/s sn m s 0.50 s =.5 m/s Insght: The ball wll arrve at the peak of ts flght at t = v sn g = s, so the ball has not et reached the peak. Its veloct at the peak wll b 0 m/s n the horzontal drecton, and ts acceleraton at that nstant wll be 9.8 m/s n the downward drecton, just as t s throughout the entre flght. 43. The shape of the path of a projectle s a parabola. 44. The range of a projectle s gven b the range equaton, ( ) mamum when the launch angle = 45. R = v g sn. In the absence of ar frcton the range s a 45. A projectle s an object whose moton s determned solel b the nfluence of gravt. In ths case, both Janet and Jennfer s bowlng balls are projectles, and each are accelerated the same amount, 9.8 m/s downward. We conclude that the acceleraton of Jennfer s bowlng ball s equal to the acceleraton of Janet s bowlng ball. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4

12 Pearson Phscs b James S. Walker 46. Pcture the Problem: Three projectles A,, and C are launched wth the same ntal speed and follow the ndcated paths. Strateg: Separatel consder the and motons of each projectle n order to answer the conceptual queston. Soluton:. (a) ecause v cos,f = v and cos decreases to zero as approaches 90, the projectle wth the largest launch angle has the smallest horzontal component of ts ntal veloct. The rankng of these projectles n order of ncreasng horzontal component of ntal veloct s A < < C. (b) The flght tme s longest for projectles that have the hghest vertcal component of the ntal veloct. Therefore, the rankng of these projectles n order of ncreasng tme of flght s C < < A. Insght: Projectle combnes a medum horzontal speed wth a medum tme of flght to acheve the longest range. Note that projectles A and C have the same range, but arrve there at dfferent tmes, wth C arrvng frst. 47. Pcture the Problem: Ths s a follow up queston for Guded Eample 4.. A golfer sends the ball over a 3.00 m-hgh tree that s 4.0 m awa. The ball passes drectl over the tree at t =.76 s. Strateg: Use the range equaton to fnd the mamum range of the golf ball n Guded Eample 4.. The ntal veloct was determned to be 3.5 m/s n Guded Eample 4.. The ball wll acheve ts mamum range f t s launched at 45.0 nstead of Soluton: The mamum range occurs when the launch angle s 45.0 : æ v ö ( 3.5 m/s) R = ç sn = sn( 45 ) = 8.6 m çè g 9.8 m s Insght: Ths mamum range s a lttle bt longer than the 7.8 m range acheved n Guded Eample 4. when the launch angle was Pcture the Problem: A snowball s thrown wth known horzontal and vertcal veloctes. Strateg: Use the vertcal veloct-tme equaton for projectles n order to fnd the tme requred for the snowball to reach ts hghest pont, at whch tme ts vertcal veloct s zero. Soluton: Solve the vertcal v,f = v, gt veloct-tme equaton for t: v,f v,f 0 8. m/s t = = g 9.8 m s = 0.84 s Insght: The snowball wll requre an addtonal 0.84 s to return to the vertcal heght from whch t was thrown, for a total tme of flght of.68 s. If t were launched over level ground, the snowball wll land ( 7.5 m/s)(.68 s) =.6 m downfeld, a dstance of about 4 feet. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4

13 Pearson Phscs b James S. Walker 49. Pcture the Problem: A softball s thrown wth known speed and drecton above a horzontal surface. Strateg: Use the poston-tme equatons for projectles launched at an angle n order to fnd the and postons of the softball at a gven tme. Soluton:. (a) The horzontal veloct remans the same throughout the flght:. The vertcal poston s affected b the acceleraton of gravt: 3. Repeat these calculatons at dfferent tmes to fnd the postons: f = + ( v cos) t = 0 + é( 8 m/s) cos 35 ù ë û ( 0.50 s) f = 7.4 m = + ( v sn q) t- gt f f = 0 + é( 8 m/s) sn 35 ù ë û( 0.50 s) - ( 9.8 m s )( 0.50 s) = 3.9 m t (s) (m) (m) (b) A plot of the softball s path s depcted above. The postons at t = 0.50 s, t =.0 s, t =.5 s, and t =.0 s are depcted wth the large crcles. Insght: If the softball were launched above level ground, t would land 3 m (0 ft) downfeld after t =.0 s. 50. Pcture the Problem: A dver runs horzontall off a dvng board (the dagram nstead shows a clff) and falls down along a parabolc arc, mantanng her horzontal veloct but ganng vertcal speed as she falls. Strateg: Frst fnd the tme t takes for the dver to enter the water b usng the vertcal poston-tme equaton for an deal projectle. Use the tme to fnd the vertcal component of the dver s veloct the nstant she enters the water. The horzontal component of the dver s veloct wll reman.85 m/s throughout the flght. The dver s speed wll be a combnaton of the vertcal and horzontal components of her veloct. Soluton:. Use the vertcal postontme equaton to fnd the landng tme:. Use the landng tme to fnd,f : 3. Fnd the fnal speed from,f and,f : = + v t gt f, 0= + 0 g t land gt land ( ) = 3.00 m t = = = 0.78 s g 9.8 m s land v v v gt ( )( ),f =, = m s 0.78 s = 7.67 m/s v v v v v ( ) ( ) =,f +,f =.85 m/s m/s = 7.89 m/s Insght: Projectle problems are often solved b frst consderng the vertcal moton, whch determnes the tme of flght and the vertcal speed, and then consderng the horzontal moton. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 3

14 Pearson Phscs b James S. Walker 5. Two vectors are equal f ther magntudes and drectons are dentcal. nspecton of the fgure ou can see that vectors A, G, and J are all equal to one another. In addton, vector I s the same as vector L. 5. (a) In order for two vectors A and to cancel each other so that A+ = 0, the must have equal magntudes and opposte drectons. We conclude that the magntude of s equal to the magntude of A. (b) The drecton of must be opposte the drecton of A. 53. No. The component and the magntude can be equal f the vector has onl a sngle component. If the vector has more than one nonzero component, however, ts magntude wll alwas be greater than ether of ts components. 54. No. If a vector has a nonzero component, the smallest magntude t can have s the magntude of the component. 55. Pcture the Problem: The base runner travels from C (home plate) to frst base, then to A (second base), then to (thrd base), and fnall back to C (home plate). The base dstance of 90 ft (as labeled n the dagram) s equvalent to 7.4 m. Strateg: The dsplacement vector Δr s the same as the poston vector r f we take home plate to be the orgn of our coordnate sstem (as t s drawn). The dsplacement vector for a runner who has just ht a double s drawn. The drecton angle s measured counter-clockwse from the postve as. Soluton:. (a) Fnd the magntude of the r = + = 7.4 m m = 38.7 m r r ( ) ( ) dsplacement vector from C to A:. Fnd the drecton angle of the vector from C to A: 3. (b) Fnd the magntude of the dsplacement vector from C to : 4. Fnd the drecton angle of the vector from C to : r tan = r ær ö - -æ7.4 mö = tan tan = = 45 èç r çè7.4 m ( ) ( ) r = r + r = 0 m m = 7.4 m r æ r ö - -æ7.4 mö tan = = tan tan 90 r = = èç r çè 0 m 3. (c) For a home run the dsplacement s zero: r = 0 m, = 0 Insght: The dsplacement s alwas zero when the object (or person) returns to ts orgnal poston. In part (c) the drecton angle s techncall undefned, as t s mpossble to dscern the drecton of a vector that has zero magntude. r 56. Pcture the Problem: The gven vector components correspond to the vector r as drawn at rght. Strateg: Use the nverse tangent functon to determne the angle. Then use the Pthagorean theorem to determne the magntude of r. Soluton:. (a) Use the nverse tangent functon to fnd the dstance angle :. (b) Use the Pthagorean theorem to determne the magntude of r : = tan æ ç - ö = -34 çè 4 or 34 below the + as r = 7 m ( 4 m) ( 9.5 m) r = r + r = + 4 m r 9.5 m Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 4

15 Pearson Phscs b James S. Walker (c) If both r and r are doubled, the = tan æ ç - ö = -34 çè 4 drecton wll reman the same but the magntude wll double: r = ( 8 m) + (- 9 m) = 34 m Insght: An vector can be resolved nto two components. The ablt to convert a vector to and from ts components s an essental skll for solvng man phscs problems. 57. Pcture the Problem: The trp takes ou toward the east frst and then toward the north. The vector s depcted at rght. Strateg: Use the Pthagorean theorem to determne the magntude and the nverse tangent functon to determne the angle. Soluton:. (a) Fnd the magntude of r : ( ) ( ) r 680 m r = 680 m m = 760 m 340 m. (b) The angle s about 30, as estmated b the sketch above m 3. (c) Use the nverse tangent functon to fnd : = tan æ ç ö = 7 north of east çè680 m Insght: Good famlart wth the trgonometrc functons sne, cosne, and tangent can help ou to quckl solve problems that nvolve vector components. 58. Pcture the Problem: The two vectors A (length 50 km) and (length 0 km) are drawn at rght. Strateg: Resolve nto ts and components to answer the questons. Soluton:. (a) Fnd : ( ) = 0 km cos 70 = 4 km 70 A. ecause the vector A ponts entrel n the drecton, we can see that A = 50 unts and that vector A has the greater component. 3. (b) Fnd : ( ) = 0 km sn 70 = 3 km 4. The vector A has no component, so t s clear that vector has the greater component, at least n terms of magntude. Alternatvel, ou could argue that has a negatve component, and zero s greater than a negatve number, and therefore conclude that vector A has the greater component. Insght: An vector can be resolved nto two components. The ablt to convert a vector to and from ts components s an essental skll for solvng man phscs problems. 59. Pcture the Problem: The four possble locatons of the treasure are labeled A,, C, and D n the fgure at rght. The poston vector for locaton A s also drawn. North s up and east s to the rght. Strateg: Use the vector components to fnd the magntude and drecton of each vector. A A.0 m.0 m A 5.0 m 5.00 m palm tree C D Soluton:. Fnd the magntude of A : ( ) ( ). Fnd the drecton (from north) of A : A =.0 m m = 9.7 m A -.0 m = tan æ ç ö = 47.7 west of north çè m 3. Fnd the magntude of : ( ) ( ) =.0 m m = 4. m Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 5

16 Pearson Phscs b James S. Walker 4. Fnd the drecton (from north) of : -.0 m = tan æ ç ö = 65.6 west of north çè m 5. Fnd the magntude of C : ( ) ( ) 6. Fnd the drecton (from north) of C : C =.0 m m = 9.7 m C -.0 m = tan æ ç ö = 47.7 east of north çè m 7. Fnd the magntude of D : ( ) ( ) 8. Fnd the drecton (from north) of D : D =.0 m m = 4. m D -.0 m = tan æ ç ö = 65.6 east of north çè m Insght: If ou ever fnd a treasure map lke ths one, ou ll be glad ou mastered vectors n phscs! 60. Pcture the Problem: The whale dves along a straght lne tlted 0.0 below horzontal for 50 m as shown n the fgure. Strateg: Resolve the whale s dsplacement vector nto horzontal and vertcal components n order to fnd ts depth r and ts horzontal travel dstance r. Soluton:. (a) The depth s gven b r : r r ( ) ( ) = sn = 50 m sn 0.0 = 5 m. (b) The horzontal travel dstance s gven b r : r r ( ) ( ) = cos = 50 m cos 0.0 = 40 m = 0.4 km Insght: Note that both answers are lmted to two sgnfcant fgures, because although 0.0 has three, 50 m has onl two sgnfcant fgures. 6. Yes. The two vectors wth equal magntudes wll sum to zero f ther drectons are opposte. 6. The two vectors A and must be perpendcular to each other. Otherwse ther magntudes would be related b the law of cosnes C = A + Acos, where s the angle between A and. 63. The vectors A and must pont n the same drecton n order for ther magntudes to sum to the magntude of C. 64. Pcture the Problem: The two vectors A (length 40.0 m) and (length 75.0 m) are drawn at rght. Strateg: Add vectors A and to fnd C= A+. Soluton:. (a) A sketch (not to scale) of the vectors and ther sum s shown at rght. The vector A s moved n the dagram to show how t adds head-to-tal to. We could also have moved so that ts tal was on the head of A and found the same result, because t does not matter n whch order that ou add vectors. Insght: A detaled analss of the vector sum reveals that C = 85.8 m and C = 43.8 m, so that C has a magntude of 96.3 m and ponts 7.0 above the as A C Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 6

17 Pearson Phscs b James S. Walker 65. Pcture the Problem: The vectors nvolved n the problem are depcted at rght. Strateg: Deduce the and components of from the nformaton gven about A and C. Use the known components to estmate the length and drecton of as well as calculate them precsel. Soluton:. (a) A sketch of the vectors s shown at the rght.. (b) The vector must have an component of 75 m so that when t s added to A the components wll cancel out. It must also have a component of 95 m because that s the length of C and A has no component to contrbute. Therefore must be longer than ether A or C and t must have an angle of greater than 90. Its length s about 0 m and t ponts at about 0 counter-clockwse from the postve as. Insght: Here the length and drecton of are determned b ts and components, whch are determned from A and C. A detaled analss reveals that has a length of m and ponts 8 counterclockwse from the + as. 66. Pcture the Problem: The vectors nvolved n the problem are depcted at rght. A Strateg: Use the vector component method of addton and subtracton to determne the components of each combnaton of A and. Once the components are known, the length and drecton of each combnaton can be determned farl easl. Soluton:. (a) Determne the components of A+. Fnd the magntude of A+ 3. (b) Determne the components of A 4. Fnd the magntude of A 5. (c) Determne the components of A 6. Fnd the magntude of A : A+ = ( ) + ( ) = ( ) 5 km 5 km 0 5 km : A ( ) ( ) + = 5 km + 5 km = 6 km : A = ( ) ( ) = ( ) : A ( ) ( ) 5 km 5 km 5 5 km = = 6 km : = ( ) ( ) = ( + ) A 5 km 5 km 0 5 km : ( ) ( ) A = 5 km + 5 km = 6 km A+ Insght: Ths problem s smplfed b the fact that A and have onl one component each, but a smlar approach wll work even wth more complcated vectors. A A A O 5 5 A A+ 67. Pcture the Problem: The vectors nvolved n the problem are depcted at rght. The control tower (CT) s at the orgn and north s up n the dagram. Strateg: Subtract vector from A usng the vector component method. Soluton:. (a) A sketch of the vectors and ther dfference s shown at rght.. (b) Subtract the components: D A ( ) ( ) ( ) ( ) = = 0 km cos km cos = 30 km Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 7

18 Pearson Phscs b James S. Walker 3. Subtract the components: D A ( ) ( ) ( ) ( ) = = 0 km sn km sn = 57 km 5 4. Fnd the magntude of D: D D D ( ) ( ) = + = 30 km + 57 km = 30 km = 3. 0 m æd ö - -æ 57 km ö 5. Fnd the drecton of D: D = tan = tan =-0+ 80= 70 or 0 north of west ç è D ç è-30 km Insght: Resolvng vectors nto components takes a lttle bt of etra effort, but the answers are more accurate b usng ths approach than b addng the vectors graphcall. Notce, however, that when our calculator returns 0 as the angle n step 5, ou must have a pcture of the vectors n our head (or on paper) to correctl determne the drecton. 68. Pcture the Problem: The vectors depctng a basketball plaer s dsplacement are shown at the rght. Strateg: Add the vectors usng the component method n order to fnd the components of the vector sum. Use the components to fnd the magntude and the drecton of the vector sum. Soluton:. (a) Make estmates from the drawng: A 45 A+ + C 0 m A+ + C= m cos m cos 30. (b) Add the vector components: ( ) ( ) ( ) ( 0.0 m) ( 0.0 m) sn 45 ( 7.0 m) sn ( 30 ) A+ + C= ( 0. m) + ( 0.64 m) + + = 0. m m = 0. m 3. Use the components to fnd the magntude: A C ( ) ( ) m 4. Use the components to fnd the angle: = tan ç æ ö =.8 çè0. m Insght: Resolvng vectors nto components takes a lttle bt of etra effort, but ou can get much more accurate answers usng ths approach than b addng the vectors graphcall. Notce, however, that when our calculator returns the angle of.8 n step 4, ou must have a pcture of the vectors n our head (or on paper) to correctl determne the drecton. 69. From our perspectve the path of the randrops wll be angled opposte the drecton ou are runnng. It wll be smlar to the case where ou are standng stll but a wnd s blowng the ran toward our face. Tlt the umbrella forward to ntercept the ran drops that would otherwse follow an angled path under our umbrella and ht our bod. 70. When salng upwnd, our speed relatve to the ar s greater than the speed of the wnd tself. If ou sal downwnd, however, ou move wth the wnd, and ts speed relatve to ou s decreased. 7. Pcture the Problem: The photo shows two arplanes flng together durng a mdar refuelng operaton. Strateg: Take note of the fact that f the two arcraft have dfferent veloctes the could not reman joned together for more than an nstant. Soluton:. (a) The arcraft beng refueled must have the same veloct as the KC-0A, or 5 m/s due east.. (b) The arcraft beng refueled must have zero veloct relatve to the KC-0A. Insght: As the arcraft beng refueled approaches the KC-0A t must have a slghtl hgher speed than the KC-0A, and then needs a slghtl lower speed n order to pull awa from the KC-0A after refuelng. A+ + C 30 C ˆ Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 8

19 Pearson Phscs b James S. Walker 7. Pcture the Problem: The vectors nvolved n ths problem are depcted at rght. Strateg: Let v = pg plane s veloct wth respect to the ground, v = ap attendant s veloct wth respect to the plane, and add the vectors accordng to the relatve moton equaton to fnd v = ag attendant s veloct wth respect to the ground. Soluton: Appl the relatve moton equaton: v = v + v = + = v ag ap pg ag = 5.3 m/s (. m/s) ( 6.5 m/s) ( 5.3 m/s) Insght: If the attendant were walkng toward the front of the plane, her speed relatve to the ground would be 7.7 m/s, slghtl faster than the arplane s speed. v ag v pg v ap 73. Pcture the Problem: The vectors nvolved n ths problem are depcted at rght. Strateg: Let v = w our veloct wth respect to the walkwa, let v = wg the walkwa s veloct wth respect to the ground, and add the vectors accordng to the relatve moton equaton to fnd v = g our veloct wth respect to the ground. Then fnd the tme t takes ou to travel the 85-m dstance. Soluton:. Fnd our veloct wth respect to the walkwa:. Appl the relatve moton equaton to fnd our veloct wth respect to the ground: v w æd ö æ85 mö = = = (.5 m/s ç ) èdt çè 68 s vg = vw + vwg =.5 m/s +. m/s = 3.45 m/s ( ) ( ) ( ) Δ 85 m 3. Now fnd the tme of travel: t = = = 5 s vg 3.45 m/s Insght: The movng walkwa slashed our tme of travel from 68 s to 5 s, a factor of.7! Note that we bent the sgnfcant fgures rules a lttle bt b not roundng v to.3 m/s. Ths helped us avod roundng error. w v w v g v wg 74. Pcture the Problem: The vectors nvolved n ths problem are depcted at rght. v ag Strateg: Let v = ra the robn s veloct wth respect to the ar, let v = ag the ar veloct wth respect to the ground, and add the vectors accordng to the relatve moton equaton to fnd v = rg the robn s veloct wth respect to the ground. Soluton:. Appl the relatve moton equaton to fnd the robn s veloct wth respect to the ground: vrg = vra + vag = m/s + 6. m/s ( ) ( ) v ra v rg. Now fnd the magntude of rg : v rg = m/s + 6. m/s = 3 m/s v ( ) ( ) Insght: You can verf that the robn s veloct relatve to the ground s 3 m/s at 7 east of north. 75. Pcture the Problem: The vectors nvolved n ths problem are depcted at rght. Strateg: Let v pf = the passenger s veloct relatve to the ferr, v = pw the passenger s veloct relatve to the water, and v = fw the ferr s veloct relatve to the water. Use the relatve moton equaton to solve for v fw. Once the components of v fw are known, ts magntude and drecton can be determned. v pw v fw N E Soluton:. Fnd the ferr s veloct relatve to the water: v = v + v v = v v pw pf fw fw pw pf 30 v pf Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 9

20 Pearson Phscs b James S. Walker. Determne the components of v fw : vfw =- é ( 4.50 m/s) sn 30 ( 4.50 m/s) cos30 ù ë + - û (.50 m/s) v =- (.5 m/s) + (.40 m/s) fw 3. Fnd the drecton of v æ v ö fw, fw : - -æ.5 m/sö = tan = tan = 43 west of north ç v ç.40 m/s çè è fw, 4. Fnd the magntude of fw v : v v v ( ) ( ) Insght: If the person were to walk even faster wth respect to the ferr, then more n the westerl drecton. fw = fw, + fw, =.5 m/s +.40 m/s = 3.9 m/s v fw would have to be shorter and pont 76. Pcture the Problem: The vectors nvolved n ths problem are depcted at rght. v ag = 36 km/h Strateg: Let v = pg the plane s veloct wth respect to the ground, v = ag veloct wth respect to the ground, and v = pa the plane s veloct wth respect to the ar. Use the relatve moton equaton, v pg = v pa + v ag to construct the rght trangle shown n the dagram, and then use trgonometr to fnd the angle. ar v p g v pa = 350 km/h Soluton:. (a) Use the nverse sne functon to fnd : æv ö - ag -æ 36 km/h ö = sn sn = ç = 5.9 east of north ç v çè pa 350 km/h è. (b) The drawng above depcts the vector dagram. 3. (c) If the plane reduces ts speed but the wnd veloct remans the same, the angle found n part (a) should be ncreased n order for the plane to contnue flng due north. Insght: If the plane s speed were to be reduced to 50 km/h, the requred angle would become Pcture the Problem: The stuaton s smlar to that depcted n the fgure at rght, ecept the boat s supposed to be a jet sk. Strateg: Place the -as perpendcular to the flow of the rver, such that the rver s flowng n the negatve -drecton. Let v = bw the jet sk s veloct relatve to the water, v = bg the jet sk s veloct relatve to the ground, and v = wg the water s veloct relatve to the ground. Use the relatve moton equaton to fnd the vector v bw, and then determne ts magntude. Soluton:. Solve for v bw : v = v + v bg bw wg v bw = v bg v wg. Fnd the components of v bg : v bg = ( 9.5 m/s) cos ( 9.5 m/s) sn 0.0 = ( 8.9 m/s) + ( 3. m/s) 3. Subtract to fnd v bw : vbw = vbg - vwg = é( 8.9 m/s) ( 3. m/s) ù ë + û -(-.8 m/s) = ( 8.9 m/s) + ( 6.0 m/s) 4. Fnd the magntude of bw v bw = 8.9 m/s m/s = m/s v : ( ) ( ) Insght: Note that the 35 angle s etraneous nformaton for ths problem. If we work backwards to fnd the angle from the components of = tan = 34, not eactl 35 due to roundng errors. v we get ( ) bw Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 0

21 Pearson Phscs b James S. Walker 78. Pcture the Problem: A projectle s launched wth known horzontal and vertcal veloctes. Strateg: The horzontal component of the veloct of a projectle remans the same throughout ts flght because there s no acceleraton n the horzontal drecton (n the absence of ar frcton). Meanwhle, the vertcal component of the veloct decreases to zero at the peak of ts flght, then ncreases to ts ntal value, onl headed downward nstead of upward. Soluton:. (a) The horzontal component v,f = v, = 5.0 m/s of the projectle s veloct never changes:. (b) The vertcal component of the projectle s veloct s zero at the hghest pont of ts flght. Insght: If the projectle were launched over level ground so that ts flght s smmetrc, at the nstant t lands the veloct of the projectle would be 5 m/s to the rght and 6 m/s downward. 79. (a) oth the scoop of ce cream and the chld have the same horzontal veloct as the horse. Relatve to the chld the scoop of ce cream falls straght downward. (b) To the parents the scoop of ce cream follows a parabolc trajector because t has an ntal horzontal component of ts veloct. 80. A projectle s an object whose moton s determned solel b the nfluence of gravt. Therefore, a projectle alwas has the same acceleraton, 9.8 m/s downward, from the nstant of launch to the nstant of landng. 8. Pcture the Problem: Three projectles A,, and C are launched wth dfferent ntal speeds and angles and follow the ndcated paths. Strateg: Separatel consder the and motons of each projectle n order to answer the conceptual queston. Soluton:. (a) Snce each projectle acheves the same mamum heght, whch s determned b the ntal vertcal veloct, we conclude that all three projectles have the same ntal vertcal veloct. That means the larger the horzontal veloct, the larger the total ntal veloct. The largest ntal speed wll therefore correspond wth the longest range. The rankng s thus A < < C.. (b) The flght tme s longest for projectles that have the hghest vertcal component of the ntal veloct. In ths case each projectle has the same mamum alttude and therefore the same ntal vertcal speed. That means the all have the same tme of flght and the rankng s thus A = = C. Insght: Projectle C travels the farthest dstance n the same amount of tme because t has the hghest speed. 8. Pcture the Problem: A projectle that s launched at an angle above horzontal follows a parabolc path. Strateg: The projectle s accelerated onl b gravt, so t mantans ts horzontal veloct whle ts vertcal veloct s reduced from a large postve value (at launch) to zero (at the peak of ts flght). Therefore the speed of the projectle at the peak of ts flght s equal to ts horzontal speed at launch. Use ths fact to determne the launch angle. Soluton: Set vpeak = v and solve for : v cos peak = v = v æ ö æ ö = = = Insght: If the launch angle were 45 the speed at the peak would be ( ) - 6 m/s cos v - cos 60 ç èv ç è m/s v cos 45 = m/s cos 45 = 8.5 m/s. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4

22 Pearson Phscs b James S. Walker 83. Pcture the Problem: The ball moves along a parabolc arc, travelng horzontall at frst and then n a more downward drecton before landng.95 m from the pont at whch t left the racket. Strateg: The ball s tme of flght can be determned from ts horzontal moton because the horzontal dstance traveled and the horzontal veloct are each known. Then fnd the dstance a ball would drop from rest durng the tme of flght. That must be the orgnal heght of the ball when t left the racket. Soluton:. Fnd the tme of flght: = v,t. Fnd the dstance the ball would fall n that amount of tme:.95 m t = = = s v 4.87 m/s, = + v t gt gt f, 0= + 0 gt ( )( ) = = 9.8 m s s = m Insght: The horzontal speed (4.87 m/s = mph) and range (.95 m = 6.40 ft) ndcate ths ball s not ht ver hard. 84. Pcture the Problem: The water of Vctora Falls follows a parabolc arc, mantanng ts horzontal veloct but ganng vertcal speed as t falls. Strateg: Fnd the vertcal speed of the water after fallng 08 m b usng the veloct-poston equaton. The horzontal veloct remans constant throughout the fall. Then fnd the magntude of the veloct from the horzontal and vertcal components. Soluton:. Use the veloctposton equaton to fnd v v,f = v, gδ = 0 ( 9.8 m/s )( 0 m 08 m) = 0 m /s :. Use the components of v to fnd the speed: v v ( ) v = + = + = 3.60 m/s 0 m /s 46. m s Insght: Projectle problems are often solved b frst consderng the vertcal moton, whch determnes the tme of flght and the vertcal speed, and then consderng the horzontal moton. 85. Pcture the Problem: The baseball s launched horzontall and follows a parabolc trajector lke that pctured at rght. Strateg: Use the equatons of moton for an deal projectle to fnd the horzontal dstance the ball traveled n the gven amount of tme, assumng the ball s horzontal veloct remaned unchanged durng the flght. Then calculate the vertcal dstance t fell durng the flght. Soluton:. (a) Fnd the horzontal v t ( )( ) dstance the ball traveled: =, = m/s 0.45 s = 9.9 m. (b) Fnd the vertcal drop durng the tme of flght: gt ( )( ) Δ = = 9.8 m/s 0.45 s = 0.99 m Insght: The drop dstance could cause the second baseman to mss. A good shortstop wll compensate b throwng the ball slghtl upwards, so that t reaches the second basement at chest level. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4

23 Pearson Phscs b James S. Walker 86. Pcture the Problem: A crow drops a clam that follows the trajector shown n the fgure at the rght. Strateg: Use the equatons of moton for an deal projectle to fnd the horzontal and vertcal veloctes of the clam after t has been dropped b the crow. Soluton:. (a) The horzontal veloct remans constant at.70 m/s.. (b) The vertcal speed ncreases due to the acceleraton of gravt: ( )( ) v,f = gt = 9.8 m/s.0 s = 0.6 m/s Insght: The speed of the crow determnes v and the acceleraton of gravt determnes v. As n most of the problems n ths chapter, the answers are accurate as long as we neglect the effects of ar frcton. 87. Pcture the Problem: A pumpkn follows a parabolc trajector after t s launched from the top of a tall tower. Strateg: Frst fnd the tme requred for the pumpkn to drop 9.0 m, then use that tme to fgure out the horzontal launch veloct (whch wll not change throughout the flght) that wll land the pumpkn on the target. Soluton:. Use the vertcal poston-tme equaton to fnd the tme of flght:. Use the horzontal postontme equaton to fnd the launch veloct: = + v t gt gt f, f = 0 ( ) ( ) f m = t = =.355 s g ( 9.8 m s ) f = + v,t f m = v, = =.6 m/s t.355 s Insght: It s often the case that the vertcal porton of a projectle s moton s used to determne tme nformaton about the trajector, and the horzontal porton s used to fnd dstance nformaton. 88. Pcture the Problem: A pumpkn s launched over level ground durng a contest and travels along a parabolc arc. Strateg: Assume the gven range corresponds to a launch angle of 45 and use the range equaton to fnd the ntal speed of the pumpkn. æ v ö æ v ö v Soluton: Solve the range equaton for v : R = sn q = sn( 45 ) = ç è g è ç g g v0 = gr = ( 9.8 m s )( 45 m) = m/s = 47 m/h Insght: The mamum range wll occur at =45 onl n the absence of ar resstance. In the presence of the atmosphere ou must launch the pumpkn at a lower angle than that n order to mamze the downrange dstance. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 3

24 Pearson Phscs b James S. Walker 89. Pcture the Problem: A dver runs horzontall off a dvng board and falls down along a parabolc arc, mantanng her horzontal veloct but ganng vertcal speed as she falls. Strateg: Use the equatons of moton for an deal projectle n order to fnd the and postons of the dver at a gven tme. Soluton:. (a) The horzontal veloct remans the same throughout the flght:. The vertcal poston s affected b the acceleraton of gravt: 3. Repeat these calculatons at dfferent tmes to fnd the postons: = + v t f, ( )( ) = m/s 0.5 s f = 0.45 m f = + v,t gt = m s 0.5 s =.7 m f ( )( ) t (s) (m) (m) (b) A plot of the dver s path s depcted above. The postons at t = 0.5 s, t = 0.50 s, and t = 0.75 s are depcted wth the large crcles. Insght: The dver wll land at tme t g ( ) ( ) = = 6.0 m 9.8 m s = 0.78 s. 90. Pcture the Problem: The stuffed anmal s trajector s depcted n the fgure at rght. Strateg: Determne the average speed of the rders on the Ferrs wheel b dvdng the crcumference of the wheel b the tme to complete a revoluton. Ths becomes the ntal speed of the stuffed anmal that s launched horzontall. Use the equatons of moton for an deal projectle, together wth the ntal speed and heght of the stuffed anmal, to determne the locaton t lands. Soluton:. (a) Fnd the crcumference of the Ferrs wheel:. Fnd the average speed of a rder: ( ) C = πr = π 5.00 m = 3.4 m C 3.46 m v = = = 0.98 m/s Δt 3.0 s 3. (b) Use the vertcal poston-tme equaton to fnd the tme of flght. The heght of the stuffed anmal when t s dropped s 5.00 m +.75 m =.75 m: = + v t gt gt f, f = 0 ( ) ( ) f 0.75 m.55 s g = t = = ( 9.8 m s ) 4. Use t to fnd the horzontal dstance: v t ( )( ) =, = 0.98 m/s.55 s =.5 m Insght: In real lfe a lght stuffed anmal wll follow a trajector that s greatl affected b ar resstance. Coprght 04 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all coprght laws as the currentl est. No porton of ths materal ma be reproduced, n an form or b an means, wthout permsson n wrtng from the publsher. 4 4