# Newton's Laws of Motion

1. Dec 13, 2009

### TNewC

1. The problem statement, all variables and given/known data
A 65kg swimmer jumps of a 10m tower.
a) Find the swimmer's velocity when hitting the water
b) The swimmer comes to a stop 2.0m below the surface. Find the net force exerted by the water.

2. Relevant equations
F=ma?
I don't know the rest, which is part of the problem

3. The attempt at a solution
I'm entirely lost, I can't figure out how to find velocity when you are given mass and distance. I just don't understand. I only got as far as being able to state the data. Please help!

2. Dec 13, 2009

### denverdoc

well, deep breath, quiet the mind, and steel your resolve.

What force is operating on the diver and what does that do his speed? Forget the numbers for a minute.

3. Dec 13, 2009

### TNewC

The force of gravity is acting on the diver, and wouldn't that increase his speed?

4. Dec 13, 2009

### denverdoc

yes, good. Do you have a number for g, the acceleration due to gravity?

5. Dec 13, 2009

### TNewC

Yes, 9.8m/s²

6. Dec 13, 2009

### denverdoc

Perfect, now what that constant tells you is for every second, the speed will increase by 9.8m/s. So At 1 sec v=9.8m/s at 4 sec=4*9.8m/s going too slow?

7. Dec 13, 2009

### TNewC

Okay, that makes sense, but I don't have a time.

8. Dec 13, 2009

### denverdoc

Bingo. But we are given height, not the time he falls. But there is some help. By using any number of approaches we find that for any body undergoing constant acceleration or deceleration as he is about to be:

2*a*x=V'^2-V^2 where d is the distance traveled while under the influence of the force.

Edit sorry, not getting the latex to work right v' is the final velocity, v the initial velocity

9. Dec 13, 2009

### denverdoc

In this case as he "falls" off the diving board, the initial velocity is zero. So you now have a way using distance to calculate V.

10. Dec 13, 2009

### TNewC

Oh! Could I use d=1/2at² to calculate time?

11. Dec 13, 2009

### denverdoc

you could, and multiply g times the t you get. This will work, the eqn above is streamlined.

12. Dec 13, 2009

### TNewC

Okay, thank you so much! I now understand what I need to do for part a, but I still don't quite get part b...can you help me with that too?

13. Dec 13, 2009

### denverdoc

sure, the process is exactly like what we just did. Only now we have to slow the diver through the depth of the water he reaches which is the displacement in this case. We know tyhe speed as he enters the water. We could do the time thing as before, but my advice would be two use tse the eqn:

a(2*d)=v^2 where d is 2 meters. Then of course F=ma allows you to complete (b)

14. Dec 13, 2009

### TNewC

Thank you so much! You have quite possibly saved my Physics grade. I understand this now!

15. Dec 13, 2009

### denverdoc

Life was tougher before the net.....I damn well recall how greek this all seemed once upon a time.