# What force must a person exert to jump 75 cm high

1. Jan 2, 2006

### Woofuls

<i>An exceptional jumper could reach a vertical height of 0.75m. What force must a 75kg person exert to achieve this height? (assume the person crouches 0.30m to jump).</i>

Okay, I know that what I'm going to have to start with:
y = viy * t + 0.5 * g * t^2

to find max height at some velocity...
(.75m-.30m) =d (10m/s^2)= g
.45m = (viy) * t = .5 * g * t^2

Right, so I have no time, but there is:
t = (vi)/g

Hrm, I still don't have (vi) or t...

My next thought is to find maximum PE...

PE = mgh

PE = 75kg * 10m/s^2 * .3m = 225J

So, I know max energy...

Work = force * mass
225J = f * 75kg, f = 500N

So, I know he can push off with a max of 500N

500N = 75kg * a
F = m * a
a = 6.66m/s^2

so I can use
6.66*t= .45m
t=.0676secs

y = viy * t + 0.5 * g * t^2
y = (6.66) * .0676 = .5 * 10 * .0676^2
about .47m, so I'm very close!

F = m * a
F = 75kg * 6.66 =500N + 750N = 1250N required

I know this is wrong, I would appreciate some help

2. Jan 2, 2006

### Staff: Mentor

Some hints on how to understand this problem:

I assume that the "vertical height" of 0.75m means that the person's center of mass rises that much above its usual position (when the person is standing). In other words, after the person leaves the ground, he rises an additional 0.75m.

I assume that the "crouches 0.30m" means that the ground only applies a force on the person while his center of mass moves a distance of 0.30m. (That's the force you are asked to find. But don't forget gravity.)

There are several ways to solve this problem. You can use kinematics: Find the person's speed the second he leaves the ground, then find his acceleration during his jump, etc. Or you can use energy methods.

Give it another shot.