Basketball and tennis ball!

For simplicity, assume that the balls are separated by a very small distance, so that the relevant bounces happen a short time apart. This assumption isn’t necessary, but it makes for a slightly cleaner solution. Just before the basketball hits the ground, both balls are moving downward with speed (using $\frac{mv^2}{2} = mgh$ )

$v=\sqrt{2gh}.............(1)$

Just after the basketball bounces off the ground, it moves upward with speed $v$ , while the tennis ball still moves downward with speed v. The relative speed is therefore $2v$ . After the balls bounce off each other, the relative speed is still $2v$ . (This is clear if you look at things in the frame of the basketball, which is essentially a brick wall). Since the upward speed of the basketball essentially stays equal to $v$ , the upward speed of the tennis ball is $2v + v = 3v$ . By conservation of energy, it will therefore rise to a height of $H = d + \frac{(3v)^2}{(2g)}$ .

But $v^2 = 2gh$ , so we have $H = d + 9h$