What is the average force exerted on a falling diver?

[BuBbLeS01]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

So the base = 1.58s
How do we calculate the Height (Fmax)?

 

[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??

 

[Doc Al]

Homework Equations

Favg = J/change in T

This makes sense. It describes average force in terms of impulse and time.

J = 1/2 * Base * Height

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?

 

[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)

 

[Doc Al]

I thought J was equal to the area under the curve (which is usually triangular shape).

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.

 

[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

 

[Doc Al]

To find the speed I just did Distance/Time so I got 4.5m/1.58s = 2.85m/s

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.)

 

[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?

 

[Doc Al]

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.

 

[BuBbLeS01]

Vf^2 = Vi^2 + 2ad

 

[BuBbLeS01]

I get a final velocity of 9.396m/s

 

[Doc Al]

I get a final velocity of 9.396m/s

Good! Use that to find the momentum.

 

[BuBbLeS01]

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

 

[Doc Al]

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?

 

[BuBbLeS01]

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