91Ó°ÊÓ

In this problem, it is considered that the level at which the child loses contact with the surface is equivalent to the ground level, that is,$\mathbf{y}=\mathbf{0}$.

Given data.

The initial speed of the second child is $u_2=4\mathrm{m}/\mathrm{s}$.

The relation from the kinematics equation is given by

${v_2}^2={u_2}^2+2gs$.

Here, $v_2$is the final velocity of the second child, the value of which is zero because, at the maximum height, the velocity is always zero, s is the maximum altitude, and g is the gravitational acceleration.

On plugging the values in the above relation,

$\begin{array}{c}{\left(0\right)}^2={\left(4\mathrm{m}/\mathrm{s}\right)}^2+2\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)s\\ s=0.81\mathrm{m}\end{array}$.

Thus,$s=0.81\mathrm{m}$ is the maximum altitude.

The height bounced by the first child is calculated as

$\begin{array}{l}{h}_1=1.5×s\\ h_1=1.5×0.81\mathrm{m}\\ h_1=1.215\mathrm{m}\end{array}$

The relation of calculating the speed is given by

${v_1}^2={u_1}^2+2g{h}_1$.

Here, $v_1$is the final velocity of the first child, the value of which is zero, and $u_1$is the initial velocity of the first child.

On plugging the values in the above relation,

$\begin{array}{c}{\left(0\right)}^2={\left(u_1\right)}^2+2\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(1.215\mathrm{m}\right)\\ u_1=4.88\mathrm{m}/\mathrm{s}\end{array}$

Thus, $u_1=4.88\mathrm{m}/\mathrm{s}$is the required speed.

The relation of calculating time is given by

$y=u_{1}t+\frac{1}{2}g{t}^2$.

Here, t is the time for which the first child was in the air.

On plugging the values in the above relation,

$\begin{array}{c}{\left(0\right)}^2={\left(4.88\mathrm{m}/\mathrm{s}\right)}^{2}t+\frac{1}{2}\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)t^2\\ t=0.996\mathrm{s}\end{array}$

Thus, $t=0.996\mathrm{s}$is the required time.