How much force or speed is needed to jump

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Discussion Overview

The discussion revolves around the physics of jumping, specifically the force and speed required to lift a given weight to a certain height. Participants explore the underlying equations and concepts related to gravitational potential energy and kinetic energy, including the variables involved in calculating jump dynamics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how fast an object needs to accelerate to jump, specifying a weight of 100 lbs or kg.
  • Another participant introduces the equation mgh = 1/2 mv², assuming a vertical jump, and seeks clarification on its components.
  • There is a request for clarification on what mgh stands for, which is subsequently defined by another participant.
  • A participant calculates the required velocity for a 10 lb mass to jump 1 foot, arriving at a velocity of 227.33 feet per second, expressing skepticism about the result.
  • Another participant points out an algebraic error in the previous calculations, suggesting that the mass does not affect the final velocity in the equation provided.
  • One participant questions the formulation of the equation used and presents a simplified version, yielding a different velocity result of 4.009 for a 1-foot jump.
  • Another participant confirms the correctness of the simplified equation but notes a mistake in the placement of the factor of 2.
  • A participant rearranges the equation to derive a velocity for a jump of approximately 2.6 m/s, using metric units.
  • There is a clarification regarding the units of velocity and acceleration, with one participant acknowledging a previous error in unit representation.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and interpretations of the equations involved. There is no consensus on the correct approach or final values, as multiple competing calculations and interpretations are presented.

Contextual Notes

Some calculations depend on specific assumptions about gravitational acceleration and the definitions of variables. There are unresolved mathematical steps and potential errors in earlier claims that participants attempt to correct or refine.

inh
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How fast does something need to accelerate in order to jump? Say a weight of 100 lbs, or kg's, etc.
 
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How high?

The equation is:

mgh = 1/2 mv²

Which is assuming that you are jumping straight up.
 
I don't have a set height, or weight really, just wondering how to determine how much force it takes to make X amont of weight move Y feet vertically.

I get the rest of the equation, but what does mgh stand for?
 
inh said:
but what does mgh stand for?
m: mass
g: gravitational acceleration near the surface of the earth. (9.8m/s^2 in SI units, about 32 feet per second^2 in imperial units).
h: the height of the jump.
 
exactly what i was looking for

thank you very much :)
 
how off am i? using a mass of 10 lbs, height of 1 foot, and g = 32.15 feet, i get 10336.23 for mgh.

v^2 = .5 * 10 * 10336.23

v = 227.33 feet per second

so to move a 10 weight 1 foot in the air it needs to have a velocity of 227 feet per second? seems a bit high...
 
inh said:
how off am i? using a mass of 10 lbs, height of 1 foot, and g = 32.15 feet, i get 10336.23 for mgh.

v^2 = .5 * 10 * 10336.23

v = 227.33 feet per second

so to move a 10 weight 1 foot in the air it needs to have a velocity of 227 feet per second? seems a bit high...
Uhh, 10*32.15*1 = 321.5 < 10336.23
 
inh said:
how off am i? using a mass of 10 lbs, height of 1 foot, and g = 32.15 feet, i get 10336.23 for mgh.

v^2 = .5 * 10 * 10336.23

v = 227.33 feet per second

so to move a 10 weight 1 foot in the air it needs to have a velocity of 227 feet per second? seems a bit high...

Uhh your algebra is a bit off. Solve for v in the equation DaleSpam posted earlier, you'll find the mass doesn't even matter.

+DaleSpam's calculation correction.
 
isnt it 10 * 32.15^2 * 1 and not 10 * 32.15 * 1 ?

also, i was simplifying the equation earlier, and i believe i got it down to v^2 = .5gh which netted 4.009 for v for a 1 foot jump. that looks better to me, how about you?
 
  • #10
inh said:
also, i was simplifying the equation earlier, and i believe i got it down to v^2 = .5gh which netted 4.009 for v for a 1 foot jump. that looks better to me, how about you?
4.009 what?

The formula you got is almost right though - you just put your factor of 2 in the wrong place ;-)
 
  • #11
\frac{1}{2}mv^2=mgh

rearranges to: h=\frac{v^2}{2g} of v=\sqrt{2gh} so to get 1 foot (which I'm going to say is approximately a third of a meter since I'm not american) and g is approximately 10m/s%2 gets me v=\sqrt{2(10)(1/3)}\approx 2.6 m/s
 
  • #12
4.0 fps^2

looks like i was off by half :) thanks for the help guys
 
  • #13
inh said:
4.0 fps^2

looks like i was off by half :) thanks for the help guys

Note the answer for velocity has units of fps, not fps^2.
 
  • #14
ah yea, sorry about that. v = 4.0 fps, the actual acceleration is fps^2, sorry about the confusion
 

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