Question

In a circus act, a 78 kg clown is shot from a cannon with an initial velocity of 19 m/s at some unknown angle above the horizontal. A short time later the clown lands in a net that is 4.4 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:

$KE_2=10712.20\ J$          

Explanation:

Given that

Mass , m = 78 kg

Initial velocity ,v= 19 m/s

Vertical height  h= 4.4 m

Now by using energy conservation

Initial kinetic energy + Initial potential energy = Final kinetic energy +Final potential energy

KE₁ + U₁ = KE₂ + U₂

Therefore

$\dfrac{1}{2}mv^2+ 0 = KE_2+mgh$

Now by putting the values in the above equation

$\dfrac{1}{2}\times 78\times 19^2+ 0 = KE_2+78\times 9.81\times 4.4$

$KE_2=\dfrac{1}{2}\times 78\times 19^2-78\times 9.81\times 4.4$

$KE_2=10712.20\ J$

Therefore the final kinetic energy will be 10712.20 J.