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A 68-kg skydiver has a speed of 52 m/s at an altitude of 670 m above the

ground. Determine the kinetic energy possessed by the skydiver, KE =

1/2mv^2*

1 Answer

8 votes

Answer:

91936J

Step-by-step explanation:

We know that kinetic energy= 1/2 mv^2

M= mass = 68 Kg

v= velocity= 52 m/s

KE=1/2 × 68 × (52)^2

KE= 1/2 × 68 × 2704

KE= 91936J

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