A steel ball is dropped from a height h = 1 m onto a steel…

A steel ball is dropped from a height $h = 1\ \mathrm{m}$ onto a steel plate and bounces. At each bounce its speed is reduced by 10 %. How long does its third fall last?

Explanation of Method

Fall times of a bouncing steel ball.

Step 1: First fall

\[ t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2}{9.81}} \approx 0.452\ \mathrm{s}, \]

\[ v_1 = g t_1. \]

Step 2: What one bounce does

After a bounce the ball leaves with 0.9 of its landing speed. Going up and coming down again takes $v/g$ each way, so every fall that follows lasts 0.9 times as long as the one before (and reaches $0.9^2 = 0.81$ of the previous height):

\[ t_2 = 0.9\,t_1, \]

\[ t_3 = 0.9\,t_2 = 0.81\,t_1. \]

Step 3: Third fall

\[ t_3 = 0.81 \cdot 0.452\ \mathrm{s}. \]

\[t_3 \approx 0.37\ \mathrm{s}\]

Check with heights

Heights: $ 1 \to 0.81 \to 0.6561\ \mathrm{m} $; $ t_3 = \sqrt{2 \cdot 0.6561 / 9.81} \approx 0.366\ \mathrm{s}\ \checkmark $

$h\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0.656$ $0.81$ $1$ $t_3 \approx 0.37\ \mathrm{s}$ $\text{1st fall}$ $\text{2nd fall}$ $\text{3rd fall}$

falls
rises after a bounce
third fall

Height of the ball against time. Each bounce keeps 90 % of the speed, so each rise and fall lasts 90 % as long as the previous one and reaches 81 % of the previous height.

Result

\[\boxed{\,t_3 \approx 0.37\ \mathrm{s}\,}\]