How long does a ball dropped from a height of $h$ bounce? The collision number for the collision between the ground and the ball is $k$. For which balls is it useful to measure $k$ with the bouncing time? (The collision number is the ratio of the mechanical energy after and before the collision.)
A ball dropped from a height of $h_{0}$ has an initial mechanical energy of $m g h_{0}$, so after the $n$-th bounce, $m g h_{n} = k^{n} \cdot m g h_{0}$. From this, the height to which the ball rises after the $n$-th bounce is:

$$
h_{n}=k^{n} h_{0}
$$

We know that the time of fall for a freely falling object from a height $h$ is $\sqrt{2 h / g}$. Thus, a ball thrown to a height $h$ is in the air for $2 \sqrt{2 h / g}$. The total bouncing time is obtained by summing the calculated times between successive bounces, so:

$$
t=\sqrt{\frac{2 h_{0}}{g}}+\sum_{n=1} 2 \sqrt{\frac{2 h_{n}}{g}}=\sqrt{\frac{2 h_{0}}{g}}+\sum_{n=1} 2 \sqrt{\frac{2 h_{0}}{g}} \cdot k^{n / 2}
$$

Using the formula for the sum of an infinite geometric series, we get:

$$
t=\sqrt{\frac{2 h_{0}}{g}}+2 \sqrt{\frac{2 h_{0}}{g}} \frac{\sqrt{k}}{1-\sqrt{k}}=\frac{1+\sqrt{k}}{1-\sqrt{k}} \cdot \sqrt{\frac{2 h_{0}}{g}}
$$

If we want to measure $k$ using the bouncing time, we can use the following formula derived from (1):

$$
k=\left(\frac{t-\sqrt{\frac{2 h_{0}}{g}}}{t+\sqrt{\frac{2 h_{0}}{g}}}\right)^{2}
$$

To measure $k$ accurately, the assumptions used in deriving the above relationships must be valid, so:

1. the effect of air resistance should be small (this requires a ball with a high average density; the effect of air resistance is also small if the ball's speed is low or if $h_{0}$ is relatively small);
2. the collision time should be short (this requires a hard ball and a hard surface). In addition, of course,
3. $t$ must be measured accurately (this requires a well-bouncing ball so that $t$ is large).

According to the above, the ideal choice for this purpose is a rubber "trick ball" or a ping-pong ball dropped from a few centimeters.