Question

In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizontal. A short time later the clown lands in a net that is 4.1 m vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

Answer 1

Answer:

$K_{2}=7302.4J$

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the conservation of total mechanical energy

$K_{1}+U_{1}=K_{2}+U_{2}$

$K_{2}=K_{1}+U_{1}-U_{2}$

$K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}$

Since $h_{1}=0$

$K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}$

$K_{2}=\frac{1}{2}(70kg)(17m/s)^2-(70kg)(9.8m/s^2)(4.1m)=7302.4J$

$K_{2}=7302.4J$