91Ó°ÊÓ

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,

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

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

The relation from the kinematics equation is

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

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.