Question

A 50-kg athlete jumps on a trampoline hitting it vertically downward with a speed of 2 m/s. If the athlete remains in contact with the trampoline for 0.1 s and the trampoline exerts an average upward force of 2500 N on her, calculate the maximum height she can reach after leaving the trampoline. A. 0.45 m B. 1.20 m C. 0.20 nm D. 0.80 nm E. 0.15 m
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Answer #1

Impulse = change in momoment

2500*0.1 = 50*(v+2)

V =3 m/s

From equation of motion

V2 - V02 = 2 a s

02 - 32 = - 2*10*h

h = 0.45 m

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