Chapter 2, Problem 64GP

To determine

The average deceleration of the person under the water

Answer to Problem 64GP

  $-19.6\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Explanation of Solution

Given:

Initial velocity of the person as it leaves the diving board $=v_o=0\frac{\text{m}}{\text{s}}$

Height of the diving board above the water surface $=H=4\text{m}$

Vertical displacement of the person as it reaches the surface of water $=\Delta y=-H=-4\text{m}$

Final velocity of the person as it touches the surface of water $=v_f$

Average acceleration above the water surface $=a=-9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Initial velocity of the person as it enter the water $=v_o^{\text{'}}=v_f$

Vertical displacement of the person under the water $=\Delta y^{\text{'}}=2\text{m}$

Average acceleration inside water $=a^{\text{'}}$

Final velocity of the person inside water $=v_f^{\text{'}}=0\frac{\text{m}}{\text{s}}$

Formula Used:

Displacement during accelerated motion of an object is given as

  $\Delta y=\frac{v_f^2-v_o^2}{2g}$

Calculation:

Consider the motion of the person above the water surface in free fall:

Displacement of the person above the surface of water is given as

  $\Delta y=\frac{v_f^2-v_o^2}{2a}$

Inserting the values

  $\begin{array}{l}-4=\frac{v_f^2-0^2}{2(-9.8)}\\ v_f=8.85\frac{\text{m}}{\text{s}}\end{array}$

Consider the motion of the person inside water:

Displacement of the person inside water is given as

  $\Delta y^{\text{'}}=\frac{v_f^{\text{'}}{}^2-v_o^{\text{'}}{}^2}{2a^{\text{'}}}$

Inserting the values

  $\begin{array}{l}2=\frac{0^2-v_f^2}{2a^{\text{'}}}\\ 2=\frac{0^2-{8.85}^2}{2a^{\text{'}}}\\ a\text{'}=-\frac{{8.85}^2}{4}\\ a\text{'}=-19.6\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Conclusion:

Therefore, average deceleration of the person under water is $-19.6\frac{\text{m}}{{\text{s}}^{\text{2}}}$