the ball traveled 188m before landing on the ground outsidethe ball park. A. asu
ID: 1764069 • Letter: T
Question
the ball traveled 188m before landing on the ground outsidethe ball park. A. asumming the ball initial velocity was 45 angle above thehorizontal and ignoring air resistance ,what did the initial speedof the ball need to be to produce such a homerun if the ball washit at a point 0.90m above ground level? assume that the ground wasperfectly flat. B. how far would the ball be above a fence 3.0m high if thefence was 116m frim home plate? C. what is the magnitide and direction of the ball's velocitywhen it reaches the fence the ball traveled 188m before landing on the ground outsidethe ball park. A. asumming the ball initial velocity was 45 angle above thehorizontal and ignoring air resistance ,what did the initial speedof the ball need to be to produce such a homerun if the ball washit at a point 0.90m above ground level? assume that the ground wasperfectly flat. B. how far would the ball be above a fence 3.0m high if thefence was 116m frim home plate? C. what is the magnitide and direction of the ball's velocitywhen it reaches the fenceExplanation / Answer
angle of projection, =45o height of point of projection, h = 0.9m initial velocity, U = ? (a) horizontal range, R = Horizontal velocity * time offlight = ( U cos ) T 188 = U * 0.707 * T U T = 265.91 .................(1) S = U t + ( 1/2)at2 during verticalmotion, -h = ( U sin ) T - ( 1/2) g T2 - 0.9 = 265.91 * 0.707 - 4.9 T2 T2 = 38.55 T = 6.21 s Initialvelocity of projection, U = 265.91 / T = 42.82 m/s (b) time taken , t = X / ( U cos ) =116 / 30.28 = 3.83 s verticaldisplacement in 't' s , Y = ( U sin ) t - ( 1/2) gt2 = 30.28 * 3.83 - 4.9 * 3.832 = 44.1m verticaldisplacement in 't' s from ground = 44.1 + 0.9 = 45 m distance from thefence, d = 45 - 3 = 42 m (c) Vy = ( U sin - gt ) = (30.28 - 9.8 * 3.83 ) = - 7.254 m/s Vx = U cos = 30.38 m/s final velocity atthe fence, V = [ (Vx)2 + (Vy)2]1/2 = 31.23 m/s angle of velocity =arc tan ( Vy / Vx) = - 13.43o ( here '-' indicate thatbelow horizontal ) T2 = 38.55 T = 6.21 s Initialvelocity of projection, U = 265.91 / T = 42.82 m/s (b) time taken , t = X / ( U cos ) =116 / 30.28 = 3.83 s verticaldisplacement in 't' s , Y = ( U sin ) t - ( 1/2) gt2 = 30.28 * 3.83 - 4.9 * 3.832 = 44.1m verticaldisplacement in 't' s from ground = 44.1 + 0.9 = 45 m distance from thefence, d = 45 - 3 = 42 m (c) Vy = ( U sin - gt ) = (30.28 - 9.8 * 3.83 ) = - 7.254 m/s Vx = U cos = 30.38 m/s final velocity atthe fence, V = [ (Vx)2 + (Vy)2]1/2 = 31.23 m/s angle of velocity =arc tan ( Vy / Vx) = - 13.43o ( here '-' indicate thatbelow horizontal )Related Questions
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