Calculating Vertical Ball Collision Speed at 100m Cliff | Physics Question

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The problem involves calculating the initial velocity required for a tennis ball thrown upwards from the bottom of a 100m cliff to be struck by a golf ball dropped from the top. Both balls experience a downward acceleration of 10 m/s². By applying kinematic equations, specifically for the tennis ball, the equation v = vi + at is used, where the final velocity at the peak is zero. The golf ball's motion is described by d = vi x t + (at²)/2, leading to a system of equations that can be solved to find the initial velocity of the tennis ball.

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A girl at the bottom of a [tex]100\ m[/tex] high cliff throws a tennis ball vertically upwards. A boy at the very top of the cliff drops a golf ball so that it hits the tennis ball while both balls are still in the air. The acceleration of both balls can be assumed to be [tex]10ms^{-2}[/tex] downwards.

With what speed is the tennis ball thrown so that the golf ball strikes it at the top of its path?

I don't know where to start with this one. Is the question missing something? I assume that both balls were both released at the same time...

Any help would be great.

Thanks
 
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for the tennis ball: v = vi + at
0 = vi -10t
(1) write an equation with t = ...

for the golf ball: d=vi x t + (at^2)/2
vi = 0 (for the golf ball) a = 10 and substitute the t from equation (1)

you'll end up with (2) d = ...

for the tennis ball: 100 - d = vi x t + (at^2)/2
a = -10, substite equation (2) for d, and substitute (1) for t

gudluck and tell me what the answer is
 

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