How do I caculate hang time?

  1. 1. The problem statement, all variables and given/known data
    Calculate the hang time of an athlete who jumps a vertical distance of 0.58 meter.

    2. Relevant equations
    all i know is that d= 0.5m, and possibly initial velocity is 0? im not sure.

    3. The attempt at a solution
    i tried using v = d/t, even though i doubted it would work.

    (this homework is due today, i really need help)
  2. jcsd
  3. Total time (hang time) = time going up + time coming down.

    and, time up = time down

    so, 2 X time down = hang time.

    for time down use the formula (yes V initial = 0, A = -9.81)

    X final = X initial + V initial (t) + 1/2A(t^2)
  4. i understand what formula to use now, but im having trouble with the math because 0.5m=(-9.81Xt^2) / 2 and i do not know how to solve for t in that, since its squared, but over a fraction and multiplying with 9.81
  5. ...well thats order of operations. you will have serious trouble passing without knowing them....

    .58m = 0 + 0 + .5 (9.81m/s^2)(t^2)

    to get t by itself

    1) add or subtract from each side (in this case that part is 0)
    2) multiply or divide
    3) take your square root

    t^2 = the sq root of (.58m / ((.5 times 9.81))

    thats the time it takes to go down. doubling it will give you your total hang time.
  6. [tex]0.58 = \frac{1}{2} \times 9.81 \times t^2[/tex]
    Divide both sides by 1/2:
    [tex]\frac{0.58}{\frac{1}{2}} = 9.81 \times t^2[/tex]
    Divide both sides by 9.81:
    [tex]\frac{0.58}{\frac{1}{2} \times 9.81} = t^2[/tex]
    Since dividing by 1/2 is the same as multiplying by 2:
    [tex]\frac{2 \times 0.58}{9.81} = t^2[/tex]
    Taking the square root:
    [tex]\sqrt{\frac{2 \times 0.58}{9.81}} = t[/tex]
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