Question

In a circus act, a 60 kgclown is shot from a cannon with an initial velocity of60 m/s at some unknown angle above the horizontal. A short time later the clown lands in a net that is 3.9 mvertically above the clown’s initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

Short answer

The kinetic energy of the clown as he lands in the net is $K{E}_f=5.4KJ$$K{E}_f=5.4kJ$.

Step-by-step solution

Given data:

Mass of clowns,$m=60kg$

Initial velocity,$v_i=16m/s$

Final velocity, $v_f=0m/s$

The height, $h=3.9m$

To understand the concept:

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.

Calculate the kinetic energy of the clown as he lands in the net

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

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

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