Chapter 2, Problem 72GP

(a)

To determine

To Find: The maximum height reached by the second child.

Answer to Problem 72GP

  $H_{\max }=1.275\text{ m}$

Explanation of Solution

Given:

Initial speed of second child, $u_2=5.0\text{ m/s}$

Acceleration of the child, $a=g=9.8{\text{ m/s}}^2$

Formula used:

Equation of motion:

  $v^2=u^2+2as$

Where,

  $v=$ Final speed.

  $u=$ Initial speed.

  $a=$ Acceleration.

  $s=$ Distance covered.

Calculation:

Let maximum height reached by the child be $H_{\max }$ .

At maximum height, final velocity of the child would be zero.

From the equation of motion:

  $\begin{array}{c}{v}_2{}^2=u_2{}^2-2g{H}_{\max }\end{array}$

   $\begin{array}{c}{0}^2={\left(5.0\right)}^2-2\times 9.8\times H_{\max }\end{array}$

  $\begin{array}{c}{H}_{\max }=\frac{{\left(5.0\right)}^2}{2\times 9.8}\end{array}$

     $\begin{array}{c}=1.275\text{ m}\end{array}$

Conclusion:

Maximum height reached by the second child: $1.275\text{ m}$

(b)

To determine

To Find: The initial speed of first child.

Answer to Problem 72GP

  $u_1=6.12\text{ m/s}$

Explanation of Solution

Given:

Height reached by first child, $h_1=1.5\times H_{\max }=1.5\times 1.275=1.91\text{ m}$

Final speed of first child, $v_1=0$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Formula used:

Equation of motion:

  $v^2=u^2+2as$

Where,

  $v=$ Final speed.

  $u=$ Initial speed.

  $a=$ Acceleration.

  $s=$ Distance covered.

Calculation:

Let the initial speed of first child be $u_1$ .

From the equation of motion:

  $\begin{array}{c}{v}_1{}^2=u_1{}^2-2g{h}_1\\ {\left(0\right)}^2=u_1{}^2-2\times 9.8\times 1.91\\ u_1=\sqrt{2\times 9.8\times 1.91}\\ =6.12\text{ m/s}\end{array}$

Conclusion:

Initial speed of first child: $6.12\text{ m/s}$

(c)

To determine

To Find: The time of first child in the air.

Answer to Problem 72GP

  $T=7.64\text{ s}$

Explanation of Solution

Given:

Initial speed of first child, $u_1=6.12\text{ m/s}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Final speed of the first child, $v_1=0$

Formula used:

First equation of motion:

  $v=u+at$

Where,

  $v=$ Final speed.

  $u=$ Initial speed.

  $a=$ Acceleration.

  $t=$ Time

Calculation:

Let time taken by first child to go up be $t_1$

From the first equation of motion:

  $\begin{array}{c}{v}_1{}^2=u_1{}^2-gt\\ 0^2={\left(6.12\right)}^2-9.8t\\ t=\frac{{\left(6.12\right)}^2}{9.8}\\ =3.82\text{ m/s}\end{array}$

Total time of child in air:

  $\begin{array}{c}T=2t\end{array}$

    $\begin{array}{c}=2\times 3.82\end{array}$

    $\begin{array}{c}=7.64\text{ s}\end{array}$

Conclusion:

Time of first child in the air: $7.64\text{ s}$