# How much force do I exert on the Earth?

#### cs44167

Homework Statement
Calculate your acceleration while jumping, the net force that acts on you while jumping, and the force you exert on Earth when you jump.
Homework Equations
w = mg

acceleration = Net force / mass

2.2 lb = 1 kg
This question required measurements which are the following:

my weight = 108 lbs
crouching distance (the distance from my regular height to where I crouch) = 90.6 cm
jump height = 60.4 cm

I first converted lb to kg, and I got 49.09 kg. I then used this value for w = mg and inputted 9.80 for g and got w = 481.09 N. I then used this force for a = net force / mass and found the acceleration to be 9.7999 m/s/s.

This answer was wrong, and I was wondering why I went wrong in my work. The answer made sense to me — the only force, neglecting air resistance, acting on someone in the middle of their jump is acceleration due to gravity.

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#### BvU

Homework Helper
Hello Cesium, !

Doesn't your result (9.7999 m/s) look a lot like $g$ ?

(Small wonder if you first do $W = mg$ and then $a = W/m$ )​

I envy your 60 cm !

Jumping has three phases: you go from crouching to right up (accelerating), then you fly (basically free fall) and when you land you crouch again to soften the landing (decelerating).

That surprises me: I agree with your reasoning and with your result. How do you know it's wrong (do you have the right answer, or just a red cross or something ?)

Perhaps - if your teacher is a computer in a strange backward country, the answer is expected to be expressed in antique units ? (Such as lbf, feet/forthnight2 or other )

#### cs44167

Hello Cesium, !

Doesn't your result (9.7999 m/s) look a lot like $g$ ?

(Small wonder if you first do $W = mg$ and then $a = W/m$ )​

I envy your 60 cm !

Jumping has three phases: you go from crouching to right up (accelerating), then you fly (basically free fall) and when you land you crouch again to soften the landing (decelerating).

That surprises me: I agree with your reasoning and with your result. How do you know it's wrong (do you have the right answer, or just a red cross or something ?)

Perhaps - if your teacher is a computer in a strange backward country, the answer is expected to be expressed in antique units ? (Such as lbf, feet/forthnight2 or other )
It’s on a website and gives a red x when the answer is incorrect. There’s never been an issue and the answer for acceleration is to be expressed in m/s/s. The only thing I was thinking was since acceleration is a vector if not having a negative sign was the issue.

We have three submissions, the first I put 9.80 m/s/s, then -9.80 m/s/s, and now I’m stuck.

#### BvU

Homework Helper
I'm afraid I am stuck as well. Doesn't help but perhaps feels a little better.

I don't suppose the website is accessible for pagans from outside ?

#### hutchphd

I'm afraid I am stuck as well. Doesn't help but perhaps feels a little better.

I don't suppose the website is accessible for pagans from outside ?
The correct answer is a maximum upward acceleration a=.67g. This is what is required to produce a jump height of 60.4cm with an acceleration stretch over a distance of 90.6cm. Then add in the weight.

#### BvU

Homework Helper
I didn't read that in the problem statement, but I can understand that it's somewhat implicitly intended (semantics?). So a jump ends in that interpretation once the object leaves the ground.

@cs44167: If your career doesnt depend on it or you have more tries left, give it a try !

@hutchphd : you seem to know the website ?

#### hutchphd

Perhaps I overstated....I think the correct answer is 1.67W....... I don't know the website. Be glad to help more tomorrow.

"How much force do I exert on the Earth?"

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