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

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To determine the speed at which the tennis ball must be thrown for the golf ball to strike it at the peak of its trajectory, both balls are assumed to be released simultaneously. The equations of motion for both balls are set up, with the tennis ball's upward motion and the golf ball's downward drop both influenced by a downward acceleration of 10 m/s². The calculations involve substituting time and distance variables to find the necessary initial velocity for the tennis ball. The discussion highlights the need for clarity on the timing and conditions of the throw and drop. Ultimately, the goal is to find the initial velocity of the tennis ball that allows for the collision at the apex of its path.
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A girl at the bottom of a 100\ m 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 10ms^{-2} 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 waht the answer is
 
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