Physics problem? Please help!?

I'm working on my exam review, and I'm confused by this:

What vertical distance can a person with 0.7 sec hang time jump?

I wrote out the values;

x=?

T=0.7

a=9.8

Vo=0

And I thought the equation was

x=VoT+1/2aT^2 , but my answer was way too big. Any help will be much appreciated!

Comments

  • Take up as positive.

    a = -9.8 (negative because gravity acceleration points down)

    Vo = ? (initial vertical velocity, the velocity in which you leave the ground. positive because its moving up)

    Vf = 0 (final vertical velocity; you stop in mid-air at the highest vertical distance)

    h = ? (height/vertical distance)

    t = 0.7

    Vf = Vo + at

    0 = Vo + (-9.8)(0.7)

    0 = Vo - 6.86

    6.86 = Vo

    h = Vot + 1/2at^2

    h = (6.86)(0.7) + 1/2(-9.8)(0.7)^2

    h = 2.401 m

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