# Force on a Falling Diver

1. Oct 21, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
A 74.0 kg diver falls from rest into a swimming pool from a height of 4.50 m. It takes 1.58 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

2. Relevant equations
Favg = J/change in T
J = 1/2 * Base * Height

3. The attempt at a solution
So the base = 1.58s
How do we calculate the Height (Fmax)?

2. Oct 21, 2007

### BuBbLeS01

I guess J also equals the change in P (which I don't understand since I thought it was the area under the curve). Is that the right equation for J??

3. Oct 21, 2007

### Staff: Mentor

This makes sense. It describes average force in terms of impulse and time.
Huh? Where does this equation come from?

J = impulse = change in momentum.

What's the diver's change in momentum? Hint: What's his initial momentum as he just hits the water?

4. Oct 21, 2007

### BuBbLeS01

I thought J was equal to the area under the curve (which is usually triangular shape).
Pi = 74kg * 0m/s
Pf = 74kg * (4.5m/158s)

5. Oct 21, 2007

### Staff: Mentor

I suppose J would equal the area of a Force vs. Time curve. But are you given any such curve in this problem? No.

No need for any curves, just find the speed of the diver as he hits the water and use that to find his momentum.

6. Oct 21, 2007

### BuBbLeS01

To find the speed I just did Distance/Time so I got 4.5m/1.58s = 2.85m/s
J = Pf-Pi
Pf = Pi
(74kg*2.85m/s) - (74kg*0m/s) = 25.96kg*m/s
Favg = J/T = 25.96/1.58s = 16.43N*s

7. Oct 21, 2007

### Staff: Mentor

That time is the time it takes him to stop once he hits the water. It's got nothing to do with the time he takes to fall the 4.5 m. (Also, realize that the speed is not constant as he falls.) You need a little kinematics to figure out the speed. (Or you can use energy methods, if that's easier for you.)

8. Oct 21, 2007

### BuBbLeS01

What kinematics equation do I use? They all have acceleration in them, so do I just use 9.8m/s^2? I thought that was an external force and we didn't include it?
Vf = Vi +at
Can I use that one?

9. Oct 21, 2007

### Staff: Mentor

Yes, the acceleration during the dive will be 9.8 m/s^2. But, since you don't have the time, that equation isn't good enough. Find one that relates velocity to acceleration and distance.

10. Oct 21, 2007

### BuBbLeS01

Vf^2 = Vi^2 + 2ad

11. Oct 21, 2007

### BuBbLeS01

I get a final velocity of 9.396m/s

12. Oct 21, 2007

### Staff: Mentor

Good! Use that to find the momentum.

13. Oct 21, 2007

### BuBbLeS01

(74kg*9.396m/s) - (74kg*0m/s) = 695.3kg*m/s
Then...
Favg = J/T = 695.3/1.58 = 440.1 N*s

14. Oct 21, 2007

### Staff: Mentor

Good. That's the average total force on the diver: 440.1 N upwards. That includes both gravity and the force of the water. If they wanted just the average force of the water on the person, what would that be?

15. Oct 21, 2007

### BuBbLeS01

Ummm...divide by 2? I don't know? Or divide by g?