91影视

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, \({\bf{y}} = {\bf{0}}\).

Given data:

The initial speed of the second child is\({u_2} = 4\;{\rm{m/s}}\).

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,

\(\begin{aligned}{\left( 0 \right)^2} = {\left( {4\;{\rm{m/s}}} \right)^2} + 2\left( { - 9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)s\\s = 0.81\;{\rm{m}}\end{aligned}\).

Thus,\(s = 0.81\;{\rm{m}}\) is the maximum altitude.

The height bounced by the first child is calculated as

\(\begin{aligned}{h_1} = 1.5 \times s\\{h_1} = 1.5 \times 0.81\;{\rm{m}}\\{h_1} = 1.215\;{\rm{m}}\end{aligned}\)

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,

\(\begin{aligned}{\left( 0 \right)^2} = {\left( {{u_1}} \right)^2} + 2\left( { - 9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.215\;{\rm{m}}} \right)\\{u_1} = 4.88\;{\rm{m/s}}\end{aligned}\).

Thus, \({u_1} = 4.88\;{\rm{m/s}}\) 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,

\(\begin{aligned}{\left( 0 \right)^2} = {\left( {4.88\;{\rm{m/s}}} \right)^2}t + \frac{1}{2}\left( { - 9.81\;{\rm{m/}}{{\rm{s}}^2}} \right){t^2}\\t = 0.996\;{\rm{s}}\end{aligned}\).

Thus, \(t = 0.996\;{\rm{s}}\) is the required time.