# Homework Help: Finding force

1. Nov 16, 2007

### missashley

1. The problem statement, all variables and given/known data

An 88 kg man drops from rest on a diving board 3.3m above the surface of the water and comes to rest 0.58 seconds after reaching the water.

Acceleration due to gravity is 9.81 m/s^s

What force does the water exert on the man? in N

2. Relevant equations

F = ma
a = v/t

3. The attempt at a solution

3.3/0.58 = 5.689655172 m/s = v

F = 88 * 5.689655172 = 500.6896552 N

F = 500.6896552 N

Did I do this right?

2. Nov 16, 2007

### mgb_phys

First you need the speed he hit the water - you got this wrong
v^2 = u^2 + 2 a h
V^2 = 2 * g * 3.3 , v = 8.05m/s

Then the accelaration (rate of slowing down) in the water
V = u + a t,
so 0 = 8.05 - a * 0.58
a = 13.8 m/s^2

Then F = ma

3. Nov 16, 2007

### Bill Foster

Don't we need to know the force the diving board exerts on the man, too? http://www.turboconnection.com/pics/smilies/hmm.gif [Broken]

Last edited by a moderator: May 3, 2017
4. Nov 16, 2007

### Bill Foster

Never mind. He doesn't drop on to the diving board; he drops off the diving board and into the water.

5. Nov 16, 2007

### Bill Foster

Start with this equation to figure the speed he's at when he hits the water:

(1) $$a \Delta{x} = \frac{1}{2}(v^2-v_0^2)$$

Then use this equation to find his acceleraction (deceleration) when he's in the water:

(2) $$x=x_0+v t+\frac{1}{2}a t^2$$

Equation 2 can be simplified into terms of only velocity and acceleration by differentiating it with respect to time, as such:

(3) $$v=v_0+at$$

Then apply the definition of force:

(4) $$F=ma$$

That should do it.

6. Nov 16, 2007

### missashley

Wouldn't a = 4.667 because 8.05 * 0.58 = a

never mind 8.05 / .58 = 13.873

Last edited: Nov 16, 2007
7. Nov 16, 2007

### missashley

How do I find the velocity?

8. Nov 16, 2007

### missashley

yay i got it right! thanks a lot!

9. Nov 16, 2007

### Bill Foster

Use the first equation. You know $$a$$:

$$a=-g=-9.8\frac{m}{s^2}$$

You know initial velocity:

$$v_0=0$$

And you know the change in distance:

$$\Delta{x}=-3.3m$$

Actually, it should be

$$\Delta{y}=-3.3m$$

since we're vertical.

Solve for $$v$$.