- #1

geoffrey159

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## Homework Statement

A 160-lb man leaps into the air from a crouching position. His center of gravity rises 1.5ft before he leaves the ground, and it then rises 3ft to the top of his leap. What power does he develop assuming that he pushes the ground with constant force?

Ans. clue: More than 1hp, less than 10hp

## Homework Equations

Power

## The Attempt at a Solution

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I have trouble understanding this problem but made an attempt at a solution.

Call h1 = 1.5ft = 0.46 m, and h2 = 3ft = 0.92m, m = 160 * 0.45 = 72kg

As far as I understand, the power the man develops is the work done on the ground divided by the time his feet are in contact with the ground.

The working forces on the ground are the weight downward, and the ground reaction force upward. Because it is assumed constant, the ground reaction force is conservative and has a potential function which is -N x height.

So the power developped is

## P = \frac{W_g}{\triangle t} = - \frac{\triangle U_g}{\triangle t} = \frac{N-mg}{\triangle t} h_1##

__Finding the time interval :__

Because all forces are constant, it is easy to find ##\triangle t## with the impulse-momentum relationship:

## (N-mg) \triangle t = m v_{jump} - 0 = m v_{jump}##

The jumping speed is found by using conservation of energy once the man is in the air, and:

##{\triangle t} = \frac{m}{N-mg}\sqrt{2g(h_2-h_1)} ##

__Finding the ground force__:

Once again, I use conservation of energy, but this time on the ground:

##E_{jump} = E_i = 0 \Rightarrow N = \frac{1}{2h_1}m{v}_{jump}^2 + mg = mg\frac{h_2}{h_1}##All together, the power is :

## P = \frac{m}{\sqrt{2} h_1} (g(h_2-h_1))^{3/2} \approx 1.42 hp##

How ugly is this ?