9A bowling ball traveling with a constant speed hits the pins at the end of a 16.5 m long bowling lane. The bowler hears the sound of the ball hitting the pins 2.80 s after the ball is released from his hands. What is the speed of the ball, assuming that the speed of sound is 340 m/s.

The speed of an object is defined as the rate of change of the distance covered by the object in unit time between any two points. The speed of an object is a scalar quantity as it depicts only the magnitude but not the direction.

Given data.

The length of the bowling lane is $l = 16.5 \mathrm{m}$.

The total time to hear the sound of the hitting is $t = 2.80 \mathrm{s}$.

The speed of the sound is $v_{\mathrm{s}}= 340 \frac{\mathrm{m}}{\mathrm{s}}$.

Assumptions.

Let the speed of the ball be $v_{\mathrm{b}}$.

The ball takes $t_1$time to reach the end of the lane, and the sound takes $t_2$time to reach the bowler from the pins that are at the end of the bowling lane.

Now, for the ball, $t_1=\frac{l}{v_{\mathrm{b}}}$, and for the sound, $t_2=\frac{l}{v_{\mathrm{s}}}$.

The total time is

$\begin{array}{c}t=t_1+t_2\\ t= \frac{l}{v_{\mathrm{b}}} + \frac{l}{v_{\mathrm{s}}}\\ \frac{l}{v_{\mathrm{b}}}=t - \frac{l}{v_{\mathrm{s}}}\\ v_{\mathrm{b}}=\frac{l}{t - \frac{l}{v_{\mathrm{s}}}}\end{array}$

Now, substituting the values,

$\begin{array}{c}{v}_{\mathrm{b}}=\frac{16.5 \mathrm{m}}{ 2.80 \mathrm{s} - \frac{16.5 \mathrm{m}}{340 \frac{\mathrm{m}}{\mathrm{s}}} }\\ v_{\mathrm{b}}=6.0 \frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Therefore, the speed of the ball is $6.0 \frac{\mathrm{m}}{\mathrm{s}}$.