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

`K_(2)=7302.4J`

Step-by-step 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)=(1)/(2)mv_(1)^(2)+mgh_(1)-mgh_(2)`

Since
`h_(1)=0`

`K_(2)=(1)/(2)mv_(1)^(2)-mgh_(2)`

`K_(2)=(1)/(2)(70kg)(17m/s)^2-(70kg)(9.8m/s^2)(4.1m)=7302.4J`

`K_(2)=7302.4J`