Calculating the Force of a Jump on the Moon

[isaacwachsman]

Homework Statement

Calculate the height of a jump on the moon vs on earth.

Relevant Equations

F=ma, vx^2 - vox^2 / 2ax = x

From my understanding, at an elementary physics level, the height at which we can jump depends on acceleration due to gravity and the initial velocity when we leave the ground. We can calculate our initial velocity by calculating acceleration and using the distance we bend down before jumping to calculate the our velocity right as our feet leave the ground. The acceleration will be the force we exert on the ground (normal force) minus (g times our mass) all divided by our mass. Please correct me if this is wrong. Keep in mind, this is for an elementary physics class, so acceleration is constant, drag force is negligible, etc. My question is this: is the force we can exert on the ground on Earth the same as the force we can exert on the ground on the moon? Would our acceleration just be that force - (1.62)(mass) all over mass? Assume we bend down the same amount before jumping on the moon as we do on earth.

 

[haruspex]

Homework Statement:: Calculate the height of a jump on the moon vs on earth.
Relevant Equations:: F=ma, vx^2 - vox^2 / 2ax = x

is the force we can exert on the ground on Earth the same as the force we can exert on the ground on the moon?

Muscles are complicated. There is a limit to how fast they can contract. But for the purpose of the question, it's a reasonable assumption.

Homework Statement:: Calculate the height of a jump on the moon vs on earth.
Relevant Equations:: F=ma, vx^2 - vox^2 / 2ax = x

Would our acceleration just be that force - (1.62)(mass) all over mass?

If 1.62 represents g on the moon, yes.

 

[isaacwachsman]

I ran through some calculations and it seems that the height one can jump on the moon is ten fold the height they can jump on the earth. I used the same mass, force, and distance from bending to calculate the jumps on Earth and the moon. Other sources say the jump on the moon should be six-fold. Is the number I calculated different simply because of the assumptions I made and the numbers I used?

 

[haruspex]

I ran through some calculations and it seems that the height one can jump on the moon is ten fold the height they can jump on the earth. I used the same mass, force, and distance from bending to calculate the jumps on Earth and the moon. Other sources say the jump on the moon should be six-fold. Is the number I calculated different simply because of the assumptions I made and the numbers I used?

Hard to say without seeing the details of your work.

 

[isaacwachsman]

Here is some of my work from the project I’m working on for school.

 

[haruspex]

Here is some of my work from the project I’m working on for school.

A couple of issues with the equation following "The person accelerated from 0 m/s to 2.80 m/s in 0.32 meters."

 

[isaacwachsman]

A couple of issues with the equation following "The person accelerated from 0 m/s to 2.80 m/s in 0.32 meters."

Could you please explain the problem?

 

[haruspex]

Could you please explain the problem?

Ok, I see now - it's just some missing parentheses that threw me. I'll keep reading.

 

[haruspex]

Ok. The reason you get such a high ratio is that you are not counting the 0.32m dip as part of the height achieved. I.e. you find the height reached rather than the overall vertical displacement.
If we add in the 0.32m we get 0.72m on Earth, 4.36m on the Moon, a ratio of about 6.

It is much better style to keep everything symbolic, only plugging in numbers at the end. It has many advantages. I would certainly give higher credit for that.

 

[isaacwachsman]

I see... but the 4.04 m I calculated is correct as far as how far off the ground the person got?

 

[haruspex]

I see... but the 4.04 m I calculated is correct as far as how far off the ground the person got?

Looks right