A baseball is placed on top of a basketball and the two are dropped. When the basketball hits the ground the baseball shoots up, going much higher than the height it had been dropped from. How can we account for this?

The masses of the two balls are in roughly a 3:1 ratio, which is actually a special case for elastic collisions, as we will see.

In the simulation the baseball is separated from the basketball so that we can analyze the different collisions (basketball with ground, baseball with basketball) separately. In reality they take place virtually simultaneously, but the net result is the same.

Part 1 - Both balls are accelerated down by gravity.

Part 2 - The basketball, with speed v down, collides with the floor. Assume this is an elastic collision, so the ball rebounds with speed v up.

Part 3 - The balls collide.

Before the collision the baseball has momentum mv down, and the basketball has 3mv up. Total momentum is 2mv up.

The total K = 0.5 mv² + 0.5 (3m)v² = 2 mv².

Assuming an elastic collision, the solution for the 3:1 mass ratio with equal-and-opposite velocities beforehand is that the baseball has velocity 2v up afterwards, and the basketball has no momentum.

The total momentum is 2mv up.
The total kinetic energy is 0.5 m (2v)² = 2 mv².
Our solution conserves both momentum and energy.

Part 4 - The baseball shoots up. If it falls from height h and acquires speed v before the collision, a speed of 2v after the collision will take it to a height of 4h! (Conservation of energy tells us that h goes as v².)