collisions

Let no. of collisions be "n".

Then
$2nH=20metres.$

$nH=10$ ...............(1)

Time of flight for a half trip

$t=\sqrt{2H/g}$

Total time $T=2nt=8sec$

$\sqrt{8n^2H/g}=8$ ...............(2)

Substituting we (1) in (2)

We get n=8

I assumed the time that it is in contact with the ground is negligable.

The average speed of the ball is known to be $\frac{20}{8}=2.5$ m/s

The average speed depends on the height in the following way: $V_{avg}=\sqrt{\frac{gH}{2}}=2.5$

This yeilds $H=1.28$ meters

For each bounce, it will travel twice the height, which is $2.56$ meters

So the number of bounces is around $\frac{20}{2.56}=7.8$ but of course it has to be a whole number.

If the ball starts at H and bounces 7 times and then falls back to the floor, it will have travelled 19.2 meters, therefore it will reach 20 meters soon after the eighth bounce, and so the answer is 8 bounces.