How much force or speed is needed to jump

[inh]

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

 

[Dale]

How high?

The equation is:

mgh = 1/2 mv²

Which is assuming that you are jumping straight up.

 

[inh]

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?

 

[Nabeshin]

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.

 

[inh]

exactly what i was looking for

thank you very much :)

 

[inh]

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...

 

[Dale]

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

 

[Nabeshin]

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.

 

[inh]

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?

 

[diazona]

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 ;-)

 

[maverick_starstrider]

$\frac{1}{2}mv^2=mgh$

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 $v=\sqrt{2(10)(1/3)}\approx 2.6 m/s$

 

[inh]

4.0 fps^2

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

 

[Nabeshin]

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.

 

[inh]

ah yea, sorry about that. v = 4.0 fps, the actual acceleration is fps^2, sorry about the confusion