PHYS Dicuion Seion 4 Projectile Motion Week 5 The Plan Thi week your group will practice analyzing projectile otion ituation. Why do we pend a whole eion on thi topic? The anwer i that projectile otion incorporate nearly all the idea and technique we ve been learning o far: Projectile otion ha nonzero acceleration, changing velocitie, and changing diplaceent. We can therefore ue all the baic kineatic definition and graph fro week one. Projectile otion occur under contant acceleration (at leat neglecting air drag). We can therefore ue the relationhip ebodied in the three kineatic equation we learned in week two. Projectile otion occur in two dienion, o all the idea about vector, coponent, and analyzing otion in different direction all apply.
PHYS Projectile Motion Week 5 DQ) Two cannon ball, each with a kg, are fired fro ground level over a flat plateau. i fired with an initial peed of 5 / at an angle of 3 degree fro the horizontal, and i fired fro the ae pot with an initial peed of / at an angle of 45 degree fro the horizontal. a. Which ball pend ore tie in the air? Check with your group eber to ee if you all agree and how any way there are to figure out the anwer to thi quetion. The tie in the air i deterined by the y coponent of the otion. Since each ball ha the ae acceleration (g9.8 / downward), the ball with the larget upward velocity will pend the ot tie in the air. You could alo ake thi concluion fro the velocitytie kineatic equation: v v + a t t y y y top v v v v ay g g y y y y ( ) Since thee cannon ball land at the ae height they were launched fro, the total tie for thee trajectorie i twice the tie to the aiu height. The uphot of all of thi i that the cannon ball with the larget upward velocity pend the ot tie in the air. Conequently, all we have to calculate i the y-coponent of the launch velocitie. 3 v A,y (5 /) in(3 )5 / 45 v B,y ( /) in(45 )4 / Since B ha the larget upward velocity coponent, pend the ot tie in the air.
PHYS Projectile Motion Week 5 b. Which ball ha the greatet horizontal peed when it land? The horizontal peed i unchanged becaue there i no -acceleration. Since the final horizontal peed equal the initial, we can jut copare the initial -coponent of their velocitie. The ball with the larget initial -coponent of velocity will have the larget final horizontal peed. 3 v A, (5 /) co(3 )7 / 45 v B, (5 /) co(45 )4 / Since ha the larget initial -coponent of velocity, land with the larget horizontal peed. c. Which ball ha the greatet vertical peed when it land? Fro the yetry of the proble (i.e. landing at the ae height they were launched at), the vertical coponent of the final velocity will be the oppoite of the vertical coponent of the initial velocity. Uing the calculation fro part a), we ee that the two vertical velocitie will be v Af -v A -5 / v Bf -v B -4 / therefore ha the greatet vertical peed when it land. 3
PHYS Projectile Motion Week 5 d. Which ball travel farther horizontally through the air? Auing you didn t calculate the flight tie in part a), we need to calculate the flight tie for each ball fro the vertical otion. Then we can inert that tie into the poition-tie equation for the horizontal otion. Start by analyzing the otion in the y- y-dienion dienion to deterine the flight tie. y y + v yt + ayt The ball tart and land at height zero and ha acceleration (-g) (auing up i ( ) ( ) + ( v inθ ) t + ( g ) t poitive). gt t v inθ gt The t olution repreent the tarting ituation. We re intereted in the later tie when the ball hit the ground. v inθ v inθ t g We can plug thi tie into the diplaceent v. tie for the -direction to find how far each ball went. The -coponent of the acceleration i zero. Here we plug in the flight tie found fro analyzing the y-direction. Now we can plug in the nuber for each ball. ditance. travel the farthet horizontal -dienion + v + a t ( ) ( coθ ) ( ) ( coθ ) + v t + t v t v inθ ( v coθ ) g ( v ) inθ coθ g ( 5 ) A in 3 co 3 557 9. 8 ( ) B in 45 co 45 477 9. 8 Reflection: Could you have olved part d by only conidering the horizontal coponent of the ball otion? Talk thi over with your group and write down your eplanation. No, the longer air tie for ball B i copenated for by a lower -coponent of velocity. We don t know ahead of tie which factor win, longer tie or larger -velocity. We actually have to calculate the range. 4
PHYS Projectile Motion Week 5 DQ) A ball i launched fro a cliff with velocity / at an angle of 5 degree above the horizontal. If the cliff i known to be 3 eter high, how far away fro the cliff bae will the ball land? a) Draw a picture of what thi otion look like. b) Talk with your group and coe up with a plan to olve thi proble. When you have filled out the ingle heet for your table, call the TA to coe over for a checkout. 5
PHYS Projectile Motion Week 5 DQ3) Conider two projectile that are fired at the ae tie. Projectile i fired fro ground level with velocity v at an angle θ, while projectile i fired at velocity v fro a height h above the ground, a hown in the diagra below. a. Which paraeter (v, θ, v, h) do you need to know to be able to deterine which projectile land firt? The tie in the air i deterined entirely by the vertical otion, naely the initial height and the initial y-velocity. We ll therefore need the height h of the econd projectile. We already know it vertical velocity coponent (zero) becaue it fired horizontally. We know the initial height of the firt projectile (zero). To find it vertical velocity coponent, we ll need the agnitude of it initial velocity v and the firing angle θ. We therefore need to know: H, v, and θ b. Find the equation for v in ter of the variable above and other contant (g,...) in the ituation where the two ball pend the ae aount of tie in the air. We know that the two projectile pend the ae aount of tie in the air. We therefore need to epre thi tie in ter of the paraeter tated in part a. We can ue the kineatic equation with the known gravitational acceleration (g downward) to do thi. We know the initial and final poition of each of the object, their initial velocitie, and their acceleration. Since we re looking for the tie fro ditance inforation, the ditance-tie kineatic equation i a good place to tart. y y + v t + a t The firt projectile i launched fro zero height with peed v at an angle of θ. We of coure get two olution for the flight tie. The t olution correpond to our launch tie, when the firt projectile alo ha a height of zero. The econd projectile i launched fro height h with peed v horizontally (i.e. with v y ) y ( ) ( ) + ( inθ ) + ( ) gt t v inθ gt t v θ v inθ t g OR in y v t g t y y + v t + a t y y ( ) ( ) + ( ) + ( ) t h t g t h g 6
PHYS Projectile Motion Week 5 DQ4) An artillery cannon i aied at an elevation of 6 degree above the horizontal. a) With what peed ut it launch a hell to hit a target k away and 5 above the cannon altitude? 6 5 b) We can find the tie to ipact (in ter of the launch peed) fro the horizontal otion of the hell. We can then ubtitute thi tie into the equation decribing the y-otion and olve for the launch peed. direction + v t + a t Write the poition v. tie for the - direction and olve for the flight tie in ter of the initial velocity. ( ) ( coθ ) ( ) ( coθ ) + v t + t v t t v coθ Write the poition v. tie for the y- direction and inert the epreion for the flight tie. Inert the nuerical paraeter fro the proble tateent. y y + v t + a t y ( inθ ) ( ) y y + v t + g t ( v ) y ( ) + ( v inθ ) g v coθ v coθ g y tanθ co θ v v y g ( tanθ y ) co θ ( 9. 8 )( ) (( ) tan ( ) )( co ) 6 5 6 9. 8 6 (( )(. ) ( ))(. ) 73 5 5 6 7
PHYS Projectile Motion Week 5 b) At thi launch peed (alo known a uzzle peed), with what velocity (peed and direction) will the hell trike the target? Now that we know the initial launch velocity of the cannonball, we can find the tie of flight fro the otion in the -direction. We know the -velocity reain contant. We can find the y-velocity at ipact fro the velocity-tie equation in the y-direction. We know the initial velocity, o we can find the tie of flight fro the otion in the - direction (ee the firt bo of part a) The -velocity reain contant, but the y- velocity change due to the acceleration due to gravity. t v coθ 5. 87 ( 6 ) co6 v v + at v coθ + ( ) t ( ) 6 co 6 63 v v + a t y y y inθ ( g ) t ( 6 ) in 6 ( 9. 8 )( 5. 87 ) v + 46. 6 The final velocity agnitude i the vector u of the - and y-coponent of the final velocity. We can ue the Pythagorean Theore to find the final peed fro the two v v + v y.. + 63 + 46 6 78 4 coponent. ( ) ( ) We can ue trigonoetry to find the direction of the final velocity 46. 6 tan θ. 74 63 θ tan. ( 74) 36. 5 below the horizontal 8
PHYS Projectile Motion Week 5 Forula Sheet Definition Poition Velocity Acceleration v d dt a dv dt d dt Contant Acceleration v v at + + v t + at ( ) v v + a Relative Motion va, B va, E + ve, B v v E, B B, E Contant and Converion g 9. 8 3 ft ile. 69 k Quadratic Forula If ± 4 a + b + c then b b ac a 9