73 kg trampoline artist jumps vertcally upward from the top of a platform w/ v = 4.8

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  • #1
A 73 kg trampoline artist jumps vertcally upward from the top of a platform with a speed of 4.8 m/s.

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 5.6×104 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 mv2 = 0.5 kx2

(73)(9.8)(2.6 + x) + 0.5(73)(4.8)2 = 0.5(5.6 x 104)(x)2


Attempt:

I got a quadratic equation of 2.8 x 104 x2 - 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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
PhanthomJay
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Assuming your math is correct, use the positive value of x. I don't know why the problem asked for a negative value. Note that you have used the term (h + x) as the height of the artist above the fully depressed position. If x was a negative value, the artist's height above the fully depressed position would be less than his/her initial height above the unstretched trampoline, which makes no sense.
 

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