# Trampolines = Fun

1. Apr 21, 2005

### ninjagowoowoo

Q:
A 55.3 kg person jumps from a window to a fire net 21.9 m below, which stretches the net 1.04 m. Assume that the net behaves like a simple spring, and calculate how much it would stretch if the same person were lying in it.

Here's what I did:

using conservation of energy to get the velocity of the person right before it hits the net. (mgh = 0.5mv^2) Once I got the velocity, I found the acceleration caused by the net (deceleration if you will) (v^2=v^2 + 2ay). Then using that acceleration, I found the force of the net on the person (F=ma) Once I found that force, I used it to find the spring constant(k) of the "spring". (F=ky). Finally, using that k, I found the displacement of the "spring" caused by the force of just the person's weight (F=kx where F=55.3kg(9.8)) and got an answer, which was wrong.

Can anyone help me?

2. Apr 21, 2005

### Berislav

You forgot to take into consideration the gravitational potential energy differance while the person was on the trampolin.

So,

$$\frac{kx^2}{2}=mg(h+x)$$

3. Apr 21, 2005

### ninjagowoowoo

hmm I don't quite understand the formula you came up with. I understand what you are saying, but perhaps you could elaborate a little more on how you came up with that formula. Thanks.

4. Apr 21, 2005

### ninjagowoowoo

oh by the way, thanks for the help.

5. Apr 22, 2005

### Berislav

Law of conservation of energy. The gravitational potential energy is converted into the energy of the spring.

$$E=\frac{kx^2}{2}$$

The person doesn't only fall 21.9 m, he also falls for an additional 1.04 m while he's streching the trampoline.

Last edited: Apr 22, 2005