Question

What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?

Answer 1

Answer:100048

Explanation:ke=1/2(74x52)^2

Answer 2

100,048

Step-by-step explanation:

K.E = 1/2 m (v)^2

K.E = 1^/2 * 74 * (52)^2

K.E = 100,048J =100.048kJ