A baseball is hit at ground level. The ball reaches its maximum height above gro

Question

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.2 s after being hit. Then 2.7 s after reaching its maximum height, the ball barely clears a fence that is 94.0 m from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

Explanation / Answer

let v(h) and v(v) be horiz. and vertical component of velocity just after hitting the ground. after 3.2 s vert. velocity = 0 So. v(v) = g*3.2 = 9.8*3.2 = 31.36 m/s a) Max. height, H = v(v)*t - 0.5 gt^2 = 31.36*3.2 - 0.5*9.8*3.2^2 = 50.18 m (ANSWER) b) In 2.7 s the ball comes down to fence height. Fall in 2.7 s, = 0.5*9.8*2.7^2 = 35.72 m Fence ht. = 50.18 - 35.72 = 14.46 m (ANSWER) c) The ball takes same time 3.2 s to reach ground as it took to go to max ht. 50.18 m. so there is 3.2 - 2.7 s = 0.5 s to reach ground. The ball reaches fence at 94 m from point of hit after 3.2+2.7 = 5.9 s Hence, v(h)*5.9 = 94 m => v(h) = 94/5.9 = 15.93 m/s In 0.5 s ball travels 15.93*0.5 = 7.96 m away from fence. (ANSWER)

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