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At t=0 a ball is thrown straight upward from the ground level at speed Vo. At the same instant, a distance D above, a ball is thrown straight downward, also at speed Vo.


Where do the balls collide in terms of D, Vo, and g?

Verified Answer :

Edufirst

1) Upward motion

y1 = Vot - g(t^2) / 2

2) Downward motion

y2 = D - [Vot  + g(t^2) / 2]

3) Collision => y1 = y2

      y1 = Vot - g(t^2) / 2
      y2 = D - [Vot + g(t^2) / 2]

Vot - g(t^2) / 2 = D - [Vot + g(t^2) / 2]

Vot - g(t^2) / 2 = D - Vot - g(t^2)/2

D = 2Vot => t = D / (2Vo)

Substitute the value of t in the equation of y1 (it is the same if you do it in the equation of y2)

y1 = Vot - g(t^2) / 2 = Vo [D/(2Vo) ] - g [D / (2Vo)]^2  / 2

y1 = D/2 - g(D^2) / 8(Vo ^2)

Answer:   y1 = D/2 - g(D^2) / 8(Vo ^2)

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