# Physics of a Long Jump

1. Aug 16, 2011

1. The problem statement, all variables and given/known data
What is the best angle to do a long jump from a 20 metre running start.

2. Relevant equations
What is the terminal velocity of the run up?
Maximum height of the jump
Horizontal and verticle components

3. The attempt at a solution
Hello there, I am doing an EEI on the physics of long jumping and I'm not exactly sure what physics are needed to figure out the best angle of the long jump.

I have video taped myself doing long jump from a 20 metre runup and a long jump. After looking at the video footage, I calculated my takeoff angle to be approximately 25 degrees to the hoirzontal.

Also I assume that my terminal velocity of my runup is also the inital velocity of my trajectory during the actual jump.

I have also taken to account the my center of mass is in my hip position. So I don't have a complete trajectory. Because at the beginning of my jump my waist is about 80cm off the ground but when I land. I tuck my legs in so it's basically or close to 0cm off the ground. Also tucking the legs in gives me more air time so I travel further.

The problem is that I don't know exactly what I'm looking for. What do I put in my introduction? what do i need to know and what I don't need to know? What am I missing?

Help is greatly appreciated... thanks guys/gals

2. Aug 16, 2011

### jaumzaum

It seems to be the same problem as mine:

If you mean, the angle required for max distance that you can get from jumping from a heigh 20m,

$Arcsin \left( \frac {Vo}{\sqrt {2Vo² + 2gH}}\right)$

But if you mean that you run 20m before and than jump (not from a given height, as before)
The angle is 45°

It can be easy proved

The height is function of time is

$h = Vo sin(\alpha)t - (1/2)gt²$

Equaling to 0

$t = \frac{ 2 Vo sin(\alpha)} {g}$ (1)

The horizontal distance in function of time is:

$x = Vo cos(\alpha)t \Rightarrow t= \frac{x}{Vo cos(\alpha)}$ (2)

If we substitute the (2) in (1), we get

$x = \frac{2Vo² sin(\alpha) cos(\alpha)}{g}$
or $\frac{2Vo² sin(2 \alpha) }{ g}$

So we have to find the max value for $sin( 2 \alpha)$
The max value for a sin is 1, so \alpha is 45º

Last edited: Aug 16, 2011
3. Aug 16, 2011

### PeterO

When you begin your jump, your centre of mass is higher than when you land - unless to land in the standing position. [you noted that]

While a "standard projectile" will achieve maximum range when fired at 45 degrees, you can probably run faster than you can do a vertical jump, so the 45 degree option will never be for you - you would always have to slow right down just to achieve the 45 degree take off.

I do suspect that your ability to run will always outstrip your ability to jump, so you will actually just run as fast as you can, then jump as high as you can.

The top long jumpers are often the best sprinters - Carl Lewis being the most recent example of that I can recall.

4. Aug 16, 2011

Because everyone is a different height, would the answer be a function?

5. Aug 16, 2011

### PeterO

A function of what? And I don't think I know.

6. Aug 16, 2011

Because everyone that will do a long jump be a different height, wouldn't their center of mass be at a different height.

For Example: a person whos center of mass is 80cm off the ground would have more of a trajectory than someone with a center of mass at 100cm off the ground? so would there be somekind of formula to solve for whatever height someone is?

7. Aug 16, 2011

### PeterO

One way to test the advantage of build, would be to go through olympic/world championship records, and find the height of the best long jumpers for the the last 10 years and see if they are all taller. You probably need the height of all competitors to see if the tall were defeating the short or vice-versa, or did it just seem random?

8. Aug 16, 2011