# Ball from a cliff

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