Question

Short answer

The obtained solutions for parts (a), (b), and (c) are \(s = 0.81\;{\rm{m}}\), \({u_1} = 4.88\;{\rm{m/s}}\), and \(t = 0.996\;{\rm{s,}}\) respectively.

Step-by-step solution

Calculation of the maximum height

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.

Calculation of the initial speed of the first child

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.

Calculation of time

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.