Chapter 3, Problem 70GP

(a)

To determine

The initial velocity of the diver.

Answer to Problem 70GP

The initial velocity of the diver is $3.42\text{m}/\text{s}$ .

Explanation of Solution

Given info:

The height of the diving board is, $h=5\text{m}$ .

The time for strike in the water is, $t=1.3\text{s}$ .

The distance from the end of the board is, $d=3\text{m}$ .

Formula Used:

The expression to find the horizontal component of the initial velocity is,

  $u_x=\frac{d}{t}$

The expression to find the vertical component of the initial velocity is,

  $u_y=\frac{\frac{1}{2}g{t}^2-h}{t}$

The expression to find the initial velocity of the diver is,

  $u=\sqrt{u_x^2+u_y^2}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_x=\frac{3\text{m}}{1.3\text{s}}\\ =2.31\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_y=\frac{\frac{1}{2}\left(9.8\text{m}/{\text{s}}^2\right){\left(1.3\text{s}\right)}^2-5\text{m}}{1.3\text{s}}\\ =2.52\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}u=\sqrt{{\left(2.31\text{m}/\text{s}\right)}^2+{\left(2.52\text{m}/\text{s}\right)}^2}\\ =3.42\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the initial velocity of the diver is $3.42\text{m}/\text{s}$ .

(b)

To determine

The maximum height of the diver.

Answer to Problem 70GP

The maximum height of the diver is $0.324\text{m}$ .

Explanation of Solution

Formula Used:

The expression to calculate the maximum height of the diver is,

  $H=\frac{u_y^2}{2g}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}H=\frac{{\left(2.52\text{m}/\text{s}\right)}^2}{2\left(9.8\text{m}/{\text{s}}^2\right)}\\ =0.324\text{m}\end{array}$

Conclusion:

Thus, the maximum height of the diver is $0.324\text{m}$ .

(c)

To determine

The final velocity of the diver.

Answer to Problem 70GP

The final velocity of the diver is $10.48\text{m}/\text{s}$ .

Explanation of Solution

Formula Used:

The expression to calculate the vertical component of the final velocity is,

  $v_y=u_y-gt$

The horizontal component of the final velocity will remain same as the horizontal component of the initial velocity.

  $v_x=u_x=2.31\text{m}/\text{s}$

The expression to calculate the final velocity is,

  $v=\sqrt{v_x^2+v_y^2}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{v}_y=\left(2.52\text{m}/\text{s}\right)-\left(9.8\text{m}/{\text{s}}^2\right)\left(1.3\text{s}\right)\\ =-10.22\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}v=\sqrt{{\left(2.31\text{}\text{m}/\text{s}\right)}^2+{\left(-10.22\text{}\text{m}/\text{s}\right)}^2}\\ =10.48\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the final velocity of the diver is $10.48\text{m}/\text{s}$ .