# Homework Help: Long jump physics

1. Jan 15, 2005

### Blashyrkh

!urgently! long jump physics

I need the physics of the long jump. What happens between the start and the landing. I'll be grateful if you could apply formulas for the acceleration, speed, flight, etc. Thanks in advance

2. Jan 15, 2005

### DaVinci

First off, anything marked as urgent normally gets ignored.

Second, you should post a whole lot more about your own thoughts and all the work you have tried to do on the problem already, or you will probably get ignored.

People around here are more than happy to help guide you in problems.... but are dead set against doing it for you.

3. Jan 15, 2005

### Blashyrkh

I've been misunderstood.
I don't want you to do it for me. I just need help with these formulas. Should I use the one for body thrown near the surface? I didn't manage to find it, and so I don't know how to calculate the distance jumped if the man runs with approximately 30 km/h and weights about 80 kilograms.

4. Jan 15, 2005

### ceptimus

The length of the jump depends on two things:

The horizontal component of the speed.

The time that the jumper can remain clear of the ground. This in turn depends on the vertical component of the speed.

For a given speed, the optimum take off angle to produce the longest jump would be 45 degrees. But real long jumpers don't achieve anywhere near this.

A complicating factor is that a long jumper has her centre of mass higher above the ground at take-off than at landing. At take-off, she will hold her arms fairly high, and stand pretty much as tall as possible. At landing, she will have her legs bent, her arms and head held low. In a sense you could say that she alters the position of her centre of mass so as to effectively jump downhill.

5. Jan 15, 2005

### DaVinci

Yeah, this one is tricky as I do not think you could model the jumper as a particle. If you could, then it would be as easy as projectile motion. But the alterate center of mass wouldnt allow that.

6. Jan 16, 2005

### futb0l

You have to view it in the center of mass frame of reference.

7. Jan 16, 2005

### Blashyrkh

So I can't use this

x=vxvxsin2a/2g

a means alpha. (sorry but I've never wrote this kind of stuff before )
v is the speed when the jumper lifts off, and a is the angle.

Last edited: Jan 16, 2005