Balls!

The height of the ball after it has bounced $n$ times with respect to the initial height is given by

$h_n = \left( \frac 34 \right)^n$

We set this equal to $0.2 = \frac 15$ and solve the resulting equation

$\begin{aligned} \left( \frac 34 \right)^n & = \frac 15 \\ \Leftrightarrow \log_{\frac 34}\left[ \left( \frac 34 \right)^n \right] & = \log_{\frac 34}\left( \frac 15 \right) \\ \Leftrightarrow n & = \log_{\frac 34}\left( \frac 15 \right) \approx 5.59 \end{aligned}$

This means, that the ball will bounce less than $0.2$ of its initial height after $\lceil 5.59 \rceil = \boxed{6}$ bounces.

1. Let’s call the original height the ball was dropped from h. After the first bounce we know it reached 75% of this height, so we can write this as 0.75h. After the next bounce it reached a height of: 0.75 x 0.75h = 0.5625h
2. The question is asking to find the number of bounces it took for the height of the ball to be less than 20% of the original height. In other words less than 0.2h.
3. After 3 bounces, the ball reached 0.75 x 0.5625h = 0.4219h. This is only 42.19% of the original height, so it is not small enough yet.
4. After 4 bounces, the ball reached 0.75 x 0.4219h = 0.3164h. This is still not small enough.
5. After 5 bounces, the ball reached 0.75 x 0.3164h = 0.2373h. Still not small enough!
6. Finally, after 6 bounces, the ball’s height was 0.75 x 0.2373h 0.1780h, or 17.8% of its original height, which is less than 20%! So the answer is 6 bounces.