Mechanical Energy: Trampoline Artist Jumps and Depresses Spring

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A 75-kg trampoline artist jumps upward at 5.0 m/s from a platform 3.0 m high. To determine his landing speed on the trampoline, energy conservation principles are applied, calculating the potential energy lost during the fall. The trampoline acts as a spring with a spring constant of 5.2x10^4 N/m, allowing for the calculation of how far the artist depresses it upon landing. The discussion focuses on the application of mechanical energy concepts to solve the problem. Understanding these principles is crucial for accurately determining the artist's speed and the spring's compression.
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A 75-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m/s. (a) how fast is he going as he lands on the trampoline 3.0 m below (Fig. 6-23)? (b) If the trampoline behaves like a spring of spring constant 5.2x10^4 N/m, how far does he depress it?
 
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I think this is a kind of homework
 

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