A 66kg trampoline artist jumps vertically upward from the top of a platform with

Question

A 66kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.1m/s. The trampoline is 3.1m below the platform. If the trampoline behaves like a spring with spring stiffness constant 6.3×104N/m, how far does he depress it? Any depression of the trampoline from equilibrium is to be taken as a negative distance.

I was correct in determining that the speed as he lands on the trampoline is 9.3m/s.

I used (1/2)mv2=(1/2)k(s)2 and solved for s, with an answer of -0.30m, but this is wrong. Please help. Thanks!

Explanation / Answer

v^2-u^2=2as u=5.1 a=-g=-9.81 m/s^2 s=-3.1 v^2-5.1^2=2*-9.81*-3.1 v=9.32 m/s mv^2/2=kx^2/2 x=0.302 m So ur answer should be correct.

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