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