91Ó°ÊÓ

Mass of clowns,$m=60kg$

Initial velocity,$v_i=16m/s$

Final velocity, $v_f=0m/s$

The height, $h=3.9m$

According to the principle of conservation of mechanical energy, the total mechanical energy of the system is conserved, i.e., energy can neither be created nor destroyed; it can be internally converted from one form to another only if the forces acting on the system are conservative in nature.

Use the concept of conservation of mechanical energy to find the kinetic energy of the clown on the net.

Formula:

$$\begin{gathered} ME=KE+PE \\ =\frac{1}{2}m{v}^2+mgh \end{gathered}$$

Here, m is he mass, v is the velocity, g is the acceleration due to gravity having a value $9.8m/s^2$, and h is the height.

From the conservation of mechanical energy, you can write

$\frac{1}{2}m{v}_f^2+mgh=\frac{1}{2}m{v}_i^2+mg{h}_0$

Where, $v_i$is the initial velocity, $v_f$is the final velocity, and $h_0$is the initial height of the clown.

Plugging the given values, you get

$$\begin{gathered} \frac{1}{2}m{v}_f^2+\left(60kg×9.8m/s^2×3.9m\right)=\frac{1}{2}\left(60kg\right){\left(16m/s\right)}^2+\left(60kg×9.8m/s^2×0m\right) \\ \frac{1}{2}m{v}_f^2=\left(7680-2293.2\right)J \\ =5386.8J \\ =5.4kJ \end{gathered}$$

Hence, the kinetic energy of the clown as he lands in the net is 5.4 kJ .