A tennis ball of mass mt is held just above a basketball of mass mb. With their

Question

A tennis ball of mass mt is held just above a basketball of mass mb. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height h and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling.

The two balls meet in an elastic collision. To what height does the tennis ball rebound? (Use any variable or symbol stated above along with the following as necessary: g.)

I thought the answer would be...

ht = ((mb+mt) * h) / mt

But apparently this answer is incorrect. Any help would be greatly appreciated!

Explanation / Answer

Well for the first one I tried that and got too big of an equation, can you help? I did tb = tennis ball and bb= basketball v=sqrt (2as) v=sqrt (19.82h) u(tb) = - sqrt (19.82h) u(bb) = sqrt (19.82h) 1/2m(tb)u(tb)^2 + 1/2m(bb)u(bb)^2 = 1/2 m(tb)v(tb)^2 + 1/2m(bb)v(bb)^2 m(tb)u(tb) + m(bb)u(bb) = m(tb)v(tb) + m(bb)v(bb) v(tb) = (u(tb) (m(tb)-m(bb) + 2(m(bb))(u(bb))/(m(tb)+m(bb)) then KE = 1/2m(tb)v(tb)^2 = PE = m(tb)gh h=v(tb)^2/19.82 so.. if I were to just put the v(tb) into the KE formula, I'd get h=h, if I were to put what the v(tb) is equal too, it'd just be a really long equation. How would I simplify more or what would I do??? :S Thanks

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