How exactly do you calculate the jumping force necesary for a 110 Lb woman to be able to jump 60 Feet in the air?

[1dnice]

Please Forgive me if this is not in the right section (I'm new here):

 

 

The fact is that alot of novels and movies out there display impossible feats while simultaniously expecting you to belive that it is realistic. I would like to learn the above because i dont want to have to put in random numbers and expect any reader of mine to have to look it up to proove if its possible. The fact is I would like to have a character that can jump at amazing heights, but I have no idea what numbers to input when describing her leg str etc.Gravity and wind resistance are two factors in the equation, but i dont even know the equation lol . Help would be much apreciated T_T

[ewmon]

This seems to be the correct sub-forum.

 

To begin with, once the woman leaves the ground, she is "ballistic" — that is, in free fall. Let's ignore the aerodynamic drag because that's complicated and inexact. The opposite of rising to 60 feet would be if she fell from 60 feet, so let's work with that.

 

Find your linear motion formulas. Displacement under constant acceleration is equation 1c in the link I gave you. The initial velocity is zero, so the equation is just s = ½at². You know gravity is 32.174 feet/second², so plug that in for a. You're dropping 60 feet, so plug that in for s.

 

Do the multiplication on the right side, then divide both sides by your result, leaving you only t² on the right side and a number on the left. Take the square root of both sides, leaving t on the right and a number (almost 2) on the left.

 

That's how long it takes to fall 60 feet or jump up to 60 feet.

 

Next find the velocity equation for constant acceleration (it's 1b in the link I gave you). Plug in your values for a and t and zero for your initial velocity (when falling 60 feet). Do the math (multiple and add) to get you velocity when you have fallen 60 feet. It should be about 62 feet/second.

 

This is the same velocity (upwards) with which you must have to reach 60 feet.

 

Now you need to know how strong she must be to accelerate herself to that velocity before leaving the ground. At best, she only has twice the length of her femur to do this in. From where her hip is one femur length below her knee to where her hip is one femur length above her knee. This is an extreme distance, but let's use it. Let's say her femur is 1¼ feet, making two femur lengths 2½ feet.

 

Let's find the equation for velocity based on acceleration and distance traveled (it's equation 1d in the link). The initial velocity (v₀) will be zero, of course. This leaves us with v = (2as)^½, more popularly known as v=√(2as). Square both sides to get rid of the square root, leaving the 2as on the right and giving you v² on the left. Plug in 2½ for s and the velocity you need for v. Do the math to get a single number on the left side and a single number times the acceleration on the right. Solve for a.

 

That's the acceleration you need throughout your pushing to get up to speed. (About 750 to 800 ft/sec²)

 

Divide that acceleration by the acceleration for gravity (32.174 ft/sec²), and this will get you the number of g's for that acceleration. The number of g's of this acceleration should be about 24.

 

The force of 1 g on this woman is 110 pounds (which you have us as her "weight"). So multiply 110 pounds by this number of g's, and that's how much force she needs to push through her 2½ foot jump to get up to speed to reach 60 feet. It should be between 2600 and 2700 pounds. Add her regular 110 pounds onto that (because if you're only pushing with 110 pounds, you're just standing there), giving you 2700 to 2800 pounds.

 

What! You mean Crouching Tiger, Hidden Dragon is totally ridiculous? Yup. So, you'd need the weight of a car on one side of a seesaw and you on the other side to get you up to speed in 2½ feet to get you 60 feet up.

 

(I did this quickly, so I hope it's right)

Edited March 2, 2012 by ewmon

[ewmon]

The quick method (by combining the equations above) is:

 

f = w × (1 + h / p)

where:

f = force required throughout the push off (lbs)

w = weight of person (lbs)

h = height to reach (ft)

p = distance to push through (ft)

[Joatmon]

Please Forgive me if this is not in the right section (I'm new here):

 

 

The fact is that alot of novels and movies out there display impossible feats while simultaniously expecting you to belive that it is realistic. I would like to learn the above because i dont want to have to put in random numbers and expect any reader of mine to have to look it up to proove if its possible. The fact is I would like to have a character that can jump at amazing heights, but I have no idea what numbers to input when describing her leg str etc.Gravity and wind resistance are two factors in the equation, but i dont even know the equation lol . Help would be much apreciated T_T

 

60 feet into the air? Realistic? About 10 times the high jump record?

It is not always necessary to be absolutely realistic to use an extraordinary fictional character in a novel. For example "Spring-heeled Jack" as in the link:- http://en.wikipedia....ing-heeled_Jack

Edited March 3, 2012 by Joatmon