Jump height difference in two different gravities

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  • #1
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Homework Statement



Gravity varies in two different places by 0.01 m/s^2. The jumper has a mass of 72 kg. What is the difference in jump height in each location? I have a couple of height measurements, but I'm pretty sure they are irrelevant because I am looking for a difference (and the problem says the difference in the heights listed is not due to gravity).

Homework Equations



I tried using F(jump)-F(weight) = ma, where a = 2h/t^2 and F(weight) = mg.
I also tried using delta x = 0.5g*t^2

The Attempt at a Solution


For both equations, I enter in the different gravities into two equations and subtract the two equations. I used 9.8 and 9.81, though I suppose I could use any two numbers varying by 0.01.
In both attempts, I come up with a difference of 0.005, but the answer checker tells me that I am wrong.
 

Answers and Replies

  • #2
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The energy of the jump is the same.
 
  • #3
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voko, I do not need energy. Energy is not involved in this particular problem. Just acceleration, force, mass.
 
  • #4
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You can't solve this with force. It is not constant.
 
  • #5
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Here is the exact wording:

Because of Mexico City's altitude the acceleration of gravity there is lower than that in
Tokyo by about 0.01 m=s2. Suppose a high-jumper has a mass of 72 kg.

(b) Assume that he is able to jump with the same initial vertical velocity in both locations, and that all other conditions are the same except for gravity. How much higher should he be able to jump in Mexico City?

Energy is at the end of this course, and I am only 1/3 of the way in! This IS supposed to be a difficult problem... Oh well!
 
  • #6
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Okay, what have you covered in the course so far?
 
  • #7
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Just Newton's first and second laws. F=ma, v=x/t, a=v/t, and some related equations.
 
  • #8
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With "Assume that he is able to jump with the same initial vertical velocity in both locations, and that all other conditions are the same except for gravity" and a = v/t, can you figure out how long it takes in both locations to reach the top point in a jump? Then, given this time, can you compute the vertical distance?
 
  • #9
Simon Bridge
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From kinematics or conservation of energy, you will find the jump height is related to initial speed by $$v^2=2gh$$ ... from there you can easily find a ratio of the heights reached in the two locations.

However, the difference in heights is not a constant ... "how much higher" seems best interpreted as the ratio ... it is not "I jumped so-many cm higher" but "I jumped so many times higher".
 

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