A rubber ball is shot straight up from the ground with speedv 0 . Simultaneously

Question

A rubber ball is shot straight up from the ground with speedv0. Simultaneously, a second rubber ball at height hdirectly above the first ball is dropped from rest.

At what height above the ground do the balls collide? Answer shouldbe a symbolic expression in terms of v0 and g.

I'm mainly unable to write a symbolic expression in terms ofv0 and g. I've written position equations for bothballs, but I feel like I need a third equation in order to createthis symbolic expression. Any help would be appreciated.

My work so far:

x is the height where the two balls collide

x = v0t - gt2/2 [ball 1 that is launched]
x = h - gt2/2 [ball 2 that is dropped]

If you notice, that implies that h = v0t. However, if Ineed to get rid of the t terms in the first equation, I will haveto write them in terms of v0 and g. Any ideas on what todo next?

Explanation / Answer

(a) Let the time elapsed beforethey meet be t.     height dropped by second ball + heightascended by first ball = h     [1/2*g*t*t] + [vo*t - 1/2*g*t*t] = vo*t =h      so, t = h/vo      so, height at which they meet = vo*t- 1/2*g*t*t = h - 1/2*g*h*h/(vo*vo) (b) time taken by the first ball to fall back toground be t.       so, v = -gt = [(-vo) -vo]          so, t = 2vo/g        For the maximum height hthe second ball must collide the first at t = 2vo/g        so, 1/2*g*t*t =1/2*g*(2vo/g)2            = h =2vo2/g (c) maximum height attained by first ball =vo2/2g = h - 1/2*g*t*t         where 0 - vo = -g*t or t = vo/g (c) maximum height attained by first ball =vo2/2g = h - 1/2*g*t*t         where 0 - vo = -g*t or t = vo/g           so, h = vo2/g (a) Let the time elapsed beforethey meet be t.     height dropped by second ball + heightascended by first ball = h     [1/2*g*t*t] + [vo*t - 1/2*g*t*t] = vo*t =h      so, t = h/vo      so, height at which they meet = vo*t- 1/2*g*t*t = h - 1/2*g*h*h/(vo*vo) (b) time taken by the first ball to fall back toground be t.       so, v = -gt = [(-vo) -vo]          so, t = 2vo/g        For the maximum height hthe second ball must collide the first at t = 2vo/g        so, 1/2*g*t*t =1/2*g*(2vo/g)2            = h =2vo2/g (c) maximum height attained by first ball =vo2/2g = h - 1/2*g*t*t         where 0 - vo = -g*t or t = vo/g           so, h = vo2/g

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