Power developed by a jumping man

• geoffrey159
In summary, the problem presents a scenario where a 160-lb man leaps into the air from a crouching position, and his center of gravity rises 1.5ft before he leaves the ground and 3ft to the top of his leap. The question asks for the power he develops assuming he pushes the ground with constant force. After attempting a solution, it is concluded that the man develops a power of approximately 1.42 hp, with a higher maximum power as he leaves the ground. The calculation involves using conservation of energy and impulse-momentum relationships. However, there is some uncertainty regarding the question and the calculation of the ground force.
geoffrey159

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

Power

The Attempt at a Solution

[/B]
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 ?

geoffrey159 said:
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 force is constant, but the speed is not, so power will vary over time.

I don't understand how you calculated the ground force, but the result looks wrong (just an issue with the last step I think).

mfb said:
The force is constant, but the speed is not, so power will vary over time.

I don't understand how you calculated the ground force, but the result looks wrong (just an issue with the last step I think).

I don't really understand the question in the problem: what does it mean to 'develop a power' ?

My first thought was that I had to find the average power during the time the man had its feet on the ground.
Because the forces are conservative, it can be written as a difference of potential energy during that time. The potential energy during that time is ## U(y) = (mg - N) y ##.

How I calculated N :
I used conservation of mechanical energy. At the beginning the man is at crouching position at rest, so ##E_0 = 0##. At height ##h_1##, as his feet leave the ground, ##E_{h_1} = \frac{1}{2} m {v}_{jump}^2 + U(h_1) ##.

But I'm not sure :) What do you think ?

geoffrey159 said:
I don't really understand the question in the problem: what does it mean to 'develop a power' ?
Power the human needs to accelerate.
The problem statement does not make clear if average or maximum power is asked for, it is probably safe to give both (once you have one, the other one is not so complicated).

geoffrey159 said:
How I calculated N :
Okay. How did the last "=" sign work? This one:
$$\frac{1}{2h_1}m{v}_{jump}^2 + mg = mg\frac{h_2}{h_1}$$

Okay, last part is obtained by replacing ##{v}_{jump}^2 ## by its value, which is given by conservation of mechanical energy between the time the man leaves the ground, and the time he is at the top of his leap. Here, only the weight is working, so ##{v}_{jump}^2 = 2g(h_2-h_1) ##.

Let me think for maximum power ...

As far as I understand the problem statement, h2=.92m is the height difference between the point where the man leaves the ground and the highest point of the jump, then you should not subtract h1.

yes I was in doubt about that too, I also tried h2 = 4.5 ft which gives an average power of 4.02 hp.

I think maximum power occurs as the man leaves the ground because power is a strictly increasing function relative to time:

## \frac{dP}{dt} = (N-mg) \frac{dv}{dt} = \frac{(N-mg)^2}{m} > 0 ##

So ## P_{max} = (N-mg) v_{jump} = (N-mg) \sqrt{2g(h_2 - h_1)} ##

geoffrey159 said:
I think maximum power occurs as the man leaves the ground because power is a strictly increasing function relative to time:
Right.
I don't see why you subtract mg from the force, it is still a force the legs have to exert.

What is the concept of power developed by a jumping man?

The concept of power developed by a jumping man refers to the amount of energy or force that a person generates while jumping. This can be measured in terms of height, distance, or speed achieved during the jump.

How is power developed by a jumping man calculated?

The power developed by a jumping man is calculated by multiplying the mass of the person by their acceleration due to gravity and the height or distance achieved during the jump. This equation is also known as the work-energy theorem.

What factors affect the power developed by a jumping man?

The power developed by a jumping man is affected by several factors, including the strength and muscle mass of the person, their technique and form while jumping, and the force they apply to the ground during takeoff.

Can power developed by a jumping man be improved?

Yes, the power developed by a jumping man can be improved through training and exercise. This includes strength training, plyometric exercises, and practicing proper jumping techniques to maximize force production.

What is the practical application of understanding power developed by a jumping man?

Understanding the power developed by a jumping man can have practical applications in sports and physical performance. It can help athletes improve their jumping abilities and overall power output, leading to better performance in activities such as basketball, volleyball, and track and field events.

• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
6K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
2K