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a) What is his speed as he lands on the trampoline, 2.6 m below his jump off point? Express your answer to 2 significant figures.

= 8.6 m/s (already figured out this to be the correct answer)

b) If the trampoline behaves like a spring with spring stiffness constant 5.6×10

^{4}N/m, how far does he depress it? Any depression of the trampoline from equilibrium is to be taken as a negative distance. Express your answer to 2 significant figures.

Equations:

mg(h + x) + 0.5 mv

^{2}= 0.5 kx

^{2}

(73)(9.8)(2.6 + x) + 0.5(73)(4.8)

^{2}= 0.5(5.6 x 10

^{4})(x)

^{2}

Attempt:

I got a quadratic equation of 2.8 x 10

^{4}x

^{2}- 775.4x - 2701

Tried solving it and got -0.35 m since the value has to be negative but it's not right so what did I do wrong?

Thanks!