91Ó°ÊÓ

The person jumps from the diving board into the water. He moves downward and stops at some distance under the water.

When the person dives, an upward force acts on him and decelerates his motion.

From the third equation of motion, the final velocity of the person can be expressed as.

$\begin{array}{c}{v}^2=u^2+2ah\\ v^2=u^2+2gh…\left(i\right)\end{array}$

Here, vis the final velocity of the person, u is the initial velocity of the person, a is the acceleration of the person, h is the height of the diving board above the water.

The given data is.

• $h=4\mathrm{m}$
  
• $a=g=9.81\mathrm{m}/{\mathrm{s}}^2$
  
• $u=0\mathrm{m}/\mathrm{s}$
  

Substitute the values of u as 0 m/s, has 4 m, and g as $9.81\mathrm{m}/{\mathrm{s}}^2$in equation (i).

$\begin{array}{c}{v}^2={\left(0\mathrm{m}/\mathrm{s}\right)}^2+2×9.81\mathrm{m}/{\mathrm{s}}^2×4\mathrm{m}\\ v=\sqrt{78.48{\mathrm{m}}^2/{\mathrm{s}}^2}\\ =8.859\mathrm{m}/\mathrm{s}\end{array}$

The speed of the person when he just strikes with the water is$8.859\mathrm{m}/\mathrm{s}$.

The distance moved by the diver below the water surface,$d=2\mathrm{m}$

From the third equation of motion, the final velocity of the person under the water can be expressed as.

role="math" localid="1643107485491" $v_1^2=u_1^2+2a_{1}d…\left(ii\right)$

Here,$v_1$is the final speed and$u_1$is the initial speed of the person when he enters the water.

The final speed of the person under the water becomes zero as he stops at a distance of 4 m under the water.

$v_1=0\mathrm{m}/\mathrm{s}$

The initial speed of the person under the water is the speed with which he strikes the water.

$u_1=v=8.859\mathrm{m}/\mathrm{s}$

Substitute the values of$v_1$as 0 m/s,$u_1$as$8.859\mathrm{m}/\mathrm{s}$, and d as$2\mathrm{m}$in equation (ii).

$\begin{array}{c}{\left(0\mathrm{m}/\mathrm{s}\right)}^2={\left(8.859\mathrm{m}/\mathrm{s}\right)}^2+2a_1×2\mathrm{m}\\ 2a×2\mathrm{m}=-{\left(8.859\mathrm{m}/\mathrm{s}\right)}^2\\ a_1=\frac{-78.482{\mathrm{m}}^2/{\mathrm{s}}^2}{4\mathrm{m}}\\ =-19.62\mathrm{m}/{\mathrm{s}}^2\end{array}$

Here, the negative sign denotes the deceleration of the person.

Thus, the average deceleration of the person is$19.62\mathrm{m}/{\mathrm{s}}^2$.