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

  1. Nov 1, 2011 #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!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 1, 2011 #2

    PhanthomJay

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

    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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