Question

A falling stone takes 0.31 s to travel past a window that is 2.2 m tall (Fig. 2-41). From what height above the top of the window did the stone fall?

Short answer

The required height is $S=1.58\mathrm{m}$.

Step-by-step solution

Step 1. Equation of motion

Average velocity of an object is defined as the average distance covered by any object in an average interval of time.

Given data.

The time is $t=0.31 \mathrm{s}$.

The height of the window is $h=2.2\mathrm{m}$.

The relation for finding the average velocity is

$v_a=\frac{h}{t}$.

On plugging the values in the above relation,

$\begin{array}{l}{v}_a=\frac{2.2\mathrm{m}}{0.31 \mathrm{s}}\\ v_a=7.09\mathrm{m}/\mathrm{s}\end{array}$

Step 2. Calculation of the initial speed

The relation of average velocity is given by

$\begin{array}{c}{v}_a=\frac{u+v}{2}\\ v=2v_a-u\end{array}$

Here, u is the initial speed.

On plugging the values in the above relation,

role="math" localid="1643022581332" $\begin{array}{l}v=2\left(7.09\mathrm{m}/\mathrm{s}\right)-u\\ v=14.18\mathrm{m}/\mathrm{s}-u\dots \left(i\right)\end{array}$

The relation from the kinematics equation is

role="math" localid="1643022606942" $\begin{array}{l}v=u+gt\\ v=u+\left(9.81 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.31 \mathrm{s}\right)\\ v=u+3.04\mathrm{m}/\mathrm{s}\dots \left(\mathrm{ii}\right)\end{array}$

Step 3. Calculation of height

On equating equations (i) and (ii),

$\begin{array}{c}14.18\mathrm{m}/\mathrm{s}-u=u+3.04\mathrm{m}/\mathrm{s}\\ u=5.57\mathrm{m}/\mathrm{s}\end{array}$

The equation of the time taken by the ball to fall from rest is

$u=v_0+gt$.

Here, $v_0$is the speed of the ball before traveling, the value of which is zero.

On plugging the values in the above relation,

$\begin{array}{c}u=0+gt\\ u=gt\\ t=\frac{u}{g}\end{array}$

The relation to calculate the height can be written as

$\begin{array}{c}S=\frac{1}{2}gt{\text{'}}^2\\ S=\frac{1}{2}g{\left(\frac{u}{g}\right)}^2\\ S=\frac{u^2}{2g}\end{array}$

On plugging the values in the above relation,

$\begin{array}{l}S=\left(\frac{{\left(5.57\mathrm{m}/\mathrm{s}\right)}^2}{2\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}\right)\\ S=1.58\mathrm{m}\end{array}$

Thus, $1.58\mathrm{m}$is the required height above the top of the window from which the stone falls.