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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?

Respuesta :

amhugueth

Answer:

[tex]K_{2}=7302.4J[/tex]

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

[tex]K_{1}+U_{1}=K_{2}+U_{2}[/tex]

[tex]K_{2}=K_{1}+U_{1}-U_{2}[/tex]

[tex]K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}[/tex]

Since [tex]h_{1}=0[/tex]

[tex]K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}[/tex]

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

[tex]K_{2}=7302.4J[/tex]

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