Trampoline Artist Jumps: Speed & Spring Stiffness Calculations

[kinematica]

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!

 

[PhanthomJay]

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.