1. The problem statement, all variables and given/known data A tennis ball of mass 57 g is held just above a basketball of mass 590 g. With their centres verticaly aligned, both are released from rest at the same moment, to fall through a distance of 1.2 m. a) Find the magnitude of the downward velocity w/ which the basketball reaches the ground b) Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. What is the velocity of the tennis ball immediately after the collision? c) Find the height to which the tennis ball rebounds. 2. Relevant equations Emech = K + U Emechi = Emechf mvi = mvf Ki = Kf 3. The attempt at a solution Well, I think I did a) right; I used Emechi = Ki + Ui = 1/2mv^2 + mgh = 1/2(0.59 g)(0)^2 + (0.59)(9.81)(1.2) = 6.94 J Emechi = Emechf 6.94 J = 1/2(0.59)v^2 + (0.59)(9.81)(0) v = 4.85 m/s But yeah. I can't seem to get b. I've tried using the formula v1i - v2i = -(v1f - v2f) but I don't know the tennis ball's initial velocity. Is there a way to find this using the whole conservation of momentum and kinetic energy spiel? Thanks.