# PROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * challenge questions

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1 PROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * hllenge questions e The ll will strike the ground 1.0 s fter it is struk. Then v x = 20 m s 1 nd v y = 0 + (9.8 m s 2 )(1.0 s) = 9.8 m s 1 The speed of the ll t 1.0 s is given y: [(20 m s 1 ) 2 + (9.8 m s 1 ) 2 ] ½ = 22.3 m s 1 Q1. A golfer prtising on rnge with n elevted tee 4.9 m ove the firwy is le to strike ll so tht it leves the lu with horizontl veloity of 20 m s 1. (Assume the elertion due to grvity is 9.80 m s 2, nd the effets of ir resistne my e ignored unless otherwise stted.) d e How long fter the ll leves the lu will it lnd on the firwy? Wht horizontl distne will the ll trvel efore striking the firwy? Wht is the elertion of the ll 0.5 s fter eing hit? Clulte the speed of the ll 0.80 s fter it leves the lu. With wht speed will the ll strike the ground? A1. x = ut + 0.5t 2 then 4.9 m = (9.8 m s 2 )t 2 nd t = 1.0 s x = (verge speed)(time) = (20 m s 1 )(1.0 s) = 20 m The elertion of the ll is onstnt t ny time during its flight, nd is equl to the elertion due to grvity = 9.8 m s 2 down d After 0.80 s, the ll hs two omponents of veloity: v x = 20 m s 1 nd v y = 0 + (9.8 m s 2 )(0.80 s) = 7.84 m s 1 The speed of the ll t 0.80 s is given y: [(20 m s 1 ) 2 + (7.84 m s 1 ) 2 ] ½ = 21.5 m s 1 Q2. A owling ll of mss 7.5 kg trvelling t 10 m s 1 rolls off horizontl tle 1.0 m high. (Assume the elertion due to grvity is 9.80 m s 2, nd the effets of ir resistne my e ignored unless otherwise stted.) Clulte the ll s horizontl veloity just s it strikes the floor. Wht is the vertil veloity of the ll s it strikes the floor? Clulte the veloity of the ll s it rehes the floor. d Wht time intervl hs elpsed etween the ll leving the tle nd striking the floor? e Clulte the horizontl distne trvelled y the ll s it flls. A2. The horizontl veloity of the ll remins onstnt nd v x = 10 m s 1. v 2 = u 2 + 2x nd v 2 y = (9.8 m s 2 )(1.0 m) nd v y = 4.4 m s 1 down v = [(10 m s 1 ) 2 + (4.43 m s 1 ) 2 ] ½ = 10.9 m s 1 t 24 to the horizontl, where the ngle is determined from tn θ = 4.43 m s 1 /10 m s 1 = nd θ = 24 d x = ut + 0.5t 2 nd 1.0 m = (9.8 m s 2 )t 2 so t = 0.45 s e Horizontl distne = (horizontl speed)(time) = (10 m s 1 )(0.45 s) = 4.5 m

2 QUESTIONS 3-8 A senior physis lss onduting reserh projet on projetile motion onstruts devie tht n lunh riket ll. The lunhing devie is designed so tht the ll n e lunhed t ground level with n initil veloity of 28 m s 1 t n ngle of 30 to the horizontl. Q3. Clulte the horizontl omponent of the veloity of the ll: initilly fter 1.0 s fter 2.0 s. A3. v x = (28 m s 1 ) os 30 = 24.2 m s 1 north nd remins onstnt throughout the flight m s 1 north 24.2 m s 1 north Q4. Clulte the vertil omponent of the veloity of the ll: initilly fter 1.0 s fter 2.0 s. A4. v y = (28 m s 1 ) sin 30 = 14 m s 1 up v y = 14 m s 1 (9.8 m s 2 )(1.0 s) = 4.2 m s 1 up The time for the ll to reh its mximum height is determined from v = u + t. Then t mximum height, the vertil veloity of the ll = 0 nd 0 = 14 m s 1 (9.8 m s 2 )t nd t = 1.43 s Therefore t t = 2.0 s the ll is 0.57 s into its downwrd flight. v y = 0 + (9.8 m s 2 )(0.57 s) = 5.6 m s 1 down Q5. At wht time will the ll reh its mximum height? Wht is the mximum height tht is hieved y the ll? Wht is the elertion of the ll t its mximum height? A5. Q6. A6. Q7. A7. The time for the ll to reh its mximum height is determined from v = u + t. Then t mximum height, the vertil veloity of the ll = 0 nd 0 = 14 m s 1 (9.8 m s 2 )t nd t = 1.43 s v 2 = u 2 + 2x then 0 = (14 m s 1 ) 2 (9.8 m s 2 )x nd x = 10 m The elertion of the ll is onstnt t ny time during its flight, nd is equl to the elertion due to grvity = 9.8 m s 2 down. At whih point in its flight will the ll experiene its minimum speed? Wht is the minimum speed of the ll during its flight? At wht time does this minimum speed our? The minimum speed will our when the vertil omponents of the ll s veloity = 0, i.e. t the mximum height. The minimum veloity of the ll during its flight ours t the mximum height, nd is equl to the horizontl omponent of the ll s veloity = 24.2 m s 1 horizontlly. The minimum speed of the ll during its flight ours t the mximum height t t = 1.43 s. At wht time fter eing lunhed will the ll return to the ground? Wht is the veloity of the ll s it strikes the ground? Clulte the horizontl rnge of the ll. The flight of the ll is symmetril. Therefore the time for it to reh the ground fter lunhing = 2(1.43 s) = 2.86 s. The flight of the ll is symmetril. Therefore the ll will strike the ground t the sme veloity s tht when it ws lunhed: 28 m s 1 t n ngle of 30 to the horizontl. Horizontl rnge = (horizontl speed)(time) = (24.2 m s 1 )(2.86 s) = 69.2 m

3 Q8*. If the effets of ir resistne were tken into ount, whih one of the following sttements would e orret? A The ll would hve trvelled greter horizontl distne efore striking the ground. B The ll would hve rehed greter mximum height. C The ll s horizontl veloity would hve een ontinully deresing. A8. C is the orret nswer. Air resistne is fore tht would e ting in the opposite diretion to the horizontl veloity of the ll, therey produing horizontl deelertion of the ll during its flight. Q9. A softll of mss 250 g is thrown with n initil veloity of 16 m s 1 t n ngle θ to the horizontl. When the ll rehes its mximum height, its kineti energy is 16 J. Wht is the mximum height hieved y the ll from its point of relese? Clulte the initil vertil veloity of the ll. Wht is the vlue of θ? d Wht is the speed of the ll fter 1.0 s? e Wht is the displement of the ll fter 1.0 s? f How long fter the ll is thrown will it return to the ground? g Clulte the horizontl distne tht the ll will trvel during its flight. A9. The initil E K of the ll = 0.5(0.250 kg)(16 m s 1 ) 2 = 32 J At its mximum height E K = 16 J, then loss in E K = 32 J 16 J = 16 J = (0.250 kg)(9.8 m s 2 )h nd mximum height h = 6.53 m v 2 = u 2 + 2x nd 0 = u 2 2(9.8 m s 2 )( m) then the initil vertil veloity u = 11.3 m s 1 up At its mximum height the veloity of the ll is equl to its horizontl omponent v x nd E K = 2 16 J = 0.5(0.250 kg)v x nd v x = m s 1 = (16 m s 1 ) os θ then os θ = nd θ = 45 d v x = (16 m s 1 ) os 45 = m s 1 v y = (9.8 m s 2 )(1.0 s) = 1.51 m s 1 then the speed of the ll t t = 1.0 s = [(11.31 m s 1 ) 2 + (1.51 m s 1 ) 2 ] ½ = 11.4 m s 1 e The horizontl displement of the ll t t = 1.0 s = (11.31 m s 1 )(1.0 s) = m, while the vertil displement t t = 1.0 s = (11.31 m s 1 )(1.0 s) 0.5(9.8 m s 2 )(1.0 s) 2 = 6.41 m The resultnt displement fter 1.0 s = [(11.31 m) 2 + (6.41 m) 2 ] ½ = 13.0 m The ngle θ of the displement from the horizontl is given y: tn θ = 6.41 m/11.31 m = nd θ = 30 f v = u + t then t mximum vertil height v y = 0 therefore 0 = m s 1 (9.8 m s 2 )t nd t = s Sine flight is symmetril, time of flight T = 2(1.154 s) = 2.31 s. g Horizontl distne = (horizontl speed)(time) = (11.31 m s 1 )(2.31 s) = 2.61 m Q10*. During trining, n eril skier tkes off from rmp tht is inlined t 40.0 to the horizontl nd lnds in pool tht is 10.0 m elow the end of the rmp. If she tkes 1.50 s to reh the highest point of her trjetory, lulte: the speed t whih she leves the rmp the mximum height ove the end of the rmp tht she rehes the time for whih she is in mid-ir. A10. Tking down s positive nd the top of the rmp s the zero position. Using the vertil omponent nd finding initil veloity y nlysing motion to mximum height: = 9.80 m s 2, t = 1.5 s, v = 0, u =? v = u + t 0 = u u v = 14.7 m s 1 ; i.e m s 1 up Trigonometry n e used to determine initil speed: v = u v /sin 40 = 14.7/0.643 = 22.9 m s 1

4 Using vertil omponent: = 9.80 m s 2, t = 1.50 s, u = 14.7 m s 1, v = 0, x =? x = ½(u + v)t = = 11.0 m Using vertil omponent: u = 14.7 m s 1, = 9.8 m s 2, x = 10 m, t =? First find finl vertil veloity (to void qudrti eqution): v =? v 2 = u 2 + 2x = ( 14.7) = 412 v = 20.3 m s 1 Now find totl flight time: t =? v = u + t 20.3 = t t = 3.57 s QUESTIONS The digrm shows projetile eing lunhed t veloity of 1.0 km s 1 t n ngle of 30 to the horizontl. Assume tht g = 9.8 m s 2 nd tht ir resistne is negligile. Q11. Wht is the time of flight of the projetile? Wht is the speed of the projetile t its mximum height? Determine the elertion of the projetile 1.0 s fter it is lunhed. A11. v = u + t 0 = [(1000 m s 1 )(sin 30 )] (9.8 ms 2 )t t = s The time of flight T (from lunh until it strikes the ground) is given y: T = 2t = 2(51.02 s) = 102 s At mximum height the projetile only hs the horizontl omponent of veloity. Then speed = the initil horizontl omponent = (1000 m s 1 )(os 30 ) = 870 m s 1 After the projetile is lunhed the only elertion it will experiene is tht due to grvity: = 9.8 m s 2 Q12. Determine the mximum vertil height tht the projetile will reh. Clulte the horizontl rnge of the projetile. Wht is the minimum speed of the projetile during its flight? A12. v 2 = u 2 + 2x 0 = [(1000 m s 1 )(sin 30 ] 2(9.8 m s 2 ) x x = 13 km x = [(1000 m s 1 )(os 30 )]( s) = 88 km The minimum speed of the projetile ours t its mximum vertil height euse the projetile only hs horizontl omponent t this height v = 870 m s 1. Q13. Clulte the vertil omponent of the projetile s veloity 1.0 s fter it is lunhed. Wht is the horizontl omponent of the projetile s veloity 1.0 s fter it is lunhed? Determine the speed of the projetile 1.0 s fter it is lunhed. A13. v = u + t = [(1000 m s 1 )(sin 30 )] (9.8 ms 2 )(1.0 s) = 490 m s 1 (1000 m s 1 )(os 30 ) = 870 m s 1 Speed = [(490 m s 1 ) 2 + (870 m s 1 ) 2 ] ½ = m s 1 Q14. A projetile is lunhed with n initil veloity of 1.5 km s 1 t n ngle θ. The trget is loted t distne of 3000 m nd on the sme level. The projetile strikes the trget fter 4.0 s. Determine the horizontl veloity of the projetile. Clulte the vlue of θ. A14. v x = x/t = (3000 m)/4.0 s = 750 m s 1 (1500 m s 1 )(os θ) = 750 m s 1 θ = 60

5 Q15. Consider projetile lunhed with initil veloity v t n ngle θ to the horizontl. Disuss the differene etween the horizontl nd vertil omponents of the projetile s veloity during the flight. Ignore ir resistne. A15. Note tht these omponents t independently of eh other. Sine we ssume tht the only fore ting on the projetile is grvity, then the horizontl omponent of the projetile s veloity will remin onstnt during its flight. The vertil omponent of the projetile s veloity will derese t onstnt rte until it rehes its mximum vertil height. It will then egin to desend. During its desent, the vertil omponent of the projetile s veloity will inrese t onstnt rte. QUESTIONS Consider projetile lunhed with n initil veloity v t n ngle θ to the horizontl. Assume tht g = 9.8 m s 2 nd tht ir resistne is negligile. Q16. Whih one or more of the following sttements re orret? A The elertion of the projetile remins onstnt during its upwrd flight. B The elertion of the projetile remins onstnt during its downwrd flight. C The elertion of the projetile inreses during its downwrd flight. D The elertion of the projetile dereses during its upwrd flight. A16. A nd B re orret sine the elertion of the projetile t ny time during its flight is equl to g = 9.8 m s 2. Q17. Whih one or more of the following sttements re orret? A B C D The vertil omponent of the projetile s veloity remins onstnt. The vertil omponent of the projetile s veloity is never zero. The horizontl omponent of the projetile s veloity remins onstnt. The horizontl omponent of the projetile s veloity is never zero. A17. C nd D re orret. Sine there re no horizontl fores ting on the projetile, the horizontl omponent of the projetile s veloity will not hnge. Sine the projetile does not tully stop during its flight, the horizontl omponent of the projetile s veloity is never zero. Q18. Whih one or more of the following sttements re orret? A The minimum speed of the projetile during its flight is zero. B The minimum speed of the projetile during its flight is equl to its initil vertil omponent of veloity. C The minimum speed of the projetile during its flight is equl to its initil horizontl omponent of veloity. D The minimum speed of the projetile during its flight is never zero. A18. C nd D re orret. The minimum speed of the projetile during its flight ours when it hs rehed its mximum vertil height when the vertil omponent of its veloity is equl to zero nd it is trvelling with its initil horizontl omponent of veloity. Q19. Whih one or more of the following sttements re orret onerning the upwrd motion of the projetile? A The vertil omponent of the projetile s veloity dereses t onstnt rte. B The vertil omponent of the projetile s veloity dereses t non-onstnt rte. C The horizontl elertion of the projetile is zero. D The vertil elertion of the projetile is zero.

6 A19. A nd C re orret. The only vertil fore ting is grvity, whih will produe onstnt deelertion of 9.8 m s 2. There re no horizontl fores ting on the projetile. QUESTIONS The digrm shows the trjetory of Vortex fter it hs een thrown with n initil speed of 10.0 m s 1. The Vortex rehes its mximum height t point Q; 4.00 m higher thn its strting height. Q20. Wht is the vlue of the ngle θ tht the initil veloity vetor mkes with the horizontl? A20. At point Q the vertil veloity = 0 Then if u = initil vertil veloity of Vortex, 0 = u 2 2(9.8 m s 2 )(4.0 m) nd u = m s 1 Sine u = (10 m s 1 ) sin θ, then θ = 62.3 Q21. Wht is the speed of the Vortex t point Q? A21. At point Q the speed of the Vortex is equl to the horizontl omponent of veloity = (10 m s 1 ) os 62.3 = 4.6 m s 1 Q22. Wht is the elertion of the Vortex t point Q? A22. The elertion of the Vortex t point Q = elertion due to grvity = 9.8 m s 2 down Q23. How fr wy is the Vortex when it rehes point R? A23. Horizontl distne X = (4.6 m s 1 )T, where T = time of flight From v = u + t, 0 = m s 1 (9.8 m s 2 )t nd t = s nd T = 2(0.903 s) = s Then X = (4.65 m s 1 )(1.806 s) = 8.4 m Q24. Two identil tennis lls X nd Y re hit horizontlly from point 2.0 m ove the ground with different initil speeds: ll X hs n initil speed of 5.0 m s 1 while ll Y hs n initil speed of 7.5 m s 1. Clulte the time it tkes for eh ll to strike the ground. Clulte the speed of ll X just efore it strikes the ground. Wht is the speed of ll Y just efore it strikes the ground? d How muh further thn ll X does ll Y trvel in the horizontl diretion efore ouning? A24. Using the formul x = ut + 0.5t 2, then 2.0 m = (9.8 m s 2 )t 2 nd t = 0.64 s For ll X: horizontl veloity = 7.5 m s 1, vertil veloity on striking ground = 0 + (9.8 m s 2 )(0.639 s) = 6.26 m s 1 then the speed of ll X just efore it strikes the ground = [(5.0 m s 1 ) 2 + (6.26 m s 1 ) 2 ] ½ = 8.0 m s 1 For ll Y: horizontl veloity = 7.5 m s 1, vertil veloity on striking ground = 0 + (9.8 m s 2 )(0.639 s) = 6.26 m s 1 then the speed of ll Y just efore it strikes the ground = [(7.5 m s 1 ) 2 + (6.26 m s 1 ) 2 ] ½ = 9.8 m s 1 d Distne = (7.5 m s m s 1 )(0.639 s) = 1.6 m

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