# Complex net force problem

1. Mar 3, 2016

### Haley Heldt

1. The problem statement, all variables and given/known data
An 82 kg man drops from rest on a diving board 3 m above the surface of the water and comes to rest .55 s after reaching the water. What is the net force on the diver as he is brought to rest

2. Relevant equations
FDt = mv F = mg v^2 = vi^2 = 2aDx Dx = 3m Dt = .33 s m = 82 kg g = 9.8m/s^2

3. The attempt at a solution
I have plunged the numbers into the equation and i found the velocity to be 7.6 m/s. Also I am pretty sure mg= 803.6 N I am just super confused on how to find the net force please help! thanks!

2. Mar 3, 2016

### Staff: Mentor

What is the guy's velocity when he hits the water. What is his velocity when he stops. What is his average deceleration? What is the average net force exerted on him by the water?

3. Mar 3, 2016

### Haley Heldt

I used the vf^2 = vi^2 + 2aDx to find that his velocity in the air is 7.6 m/s and I used the formula FDt = mv to calculate the force exerted by the water which I believe is 1133.1N . I know that he starts at rest and he eventually ends at rest however I don't know his average net force or deceleration. Sorry!

4. Mar 3, 2016

### haruspex

In the equation $\int F.dt=m\Delta v$, why should F be specifically the force from the water? What other forces act?

5. Mar 3, 2016

### Staff: Mentor

If you're traveling 7.6 m/s, and 0.55 sec later, your speed is 0, what is your average acceleration?

6. Mar 3, 2016

### haruspex

The way I read post #3, Haley is past that point.
Indeed, has the answer but does not realise it.

7. Mar 3, 2016

### Staff: Mentor

Post #3: "I don't know his average net force or deceleration"

8. Mar 3, 2016

Yes, but:

9. Mar 3, 2016

### Staff: Mentor

I see what you're saying, but she really ought to know how acceleration is defined and how net force is defined (and why). It seems to me that she luckily pulled the impulse equation out of the air without really understanding what is happening at a fundamental level.

Chet