Question: A baseball is hit at ground level. The ball is observed to reach its m

Question

Question: A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 s after being hit. And 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 320 ft from where it was hit. If the ground is level, what horizontal distance beyond the fence will the ball strike the ground?

Answer: 29 ft

Explanation:

S = (initial velocity)(t) + (1/2)(a)(t^2)

320 = (initial velocity)(5.5) + (1/2)(0)(t^2)

320 = (initial velocity)(5.5)

initial velocity = 58.18

S = (58.18)(0.5) = 29.09

Can someone explain why the acceleration is 0 in this case??

Explanation / Answer

here

given t = 3 + 2.5 = 5.5 sec

S = u X (t) + (1/2) X (a) X (t2)

320 = u X (5.5) + (1/2) X (0) X (t2)

320 = u X (5.5)

u = 320 / 5.5

u = 58.18 ft/s

s = u X t

s = 58.18 X 0.5

s = 29.09 ft

here in this it reaches to the maximum height

and

horizontal distance the accelearation is zero

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