# Sky diving problem

1. Oct 6, 2005

(a) What is the acceleration of two falling sky divers (mass 102.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?
________ m/s2 (downward)

(b) After popping open the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? See Fig. 4-34.
_____ N (upward)

i need help on starting this problem...like what equation(s) would i use??

thanks so much!

2. Oct 6, 2005

### Tom Mattson

Staff Emeritus
Instead of immediately looking for an equation, try thinking about the problem.

If the sky divers are subjected to an upward force equal to one-fourth of their total weight, then what is the net force acting on them?

Once you determine that, then think about what law of physics relates net force (which you will have determined by this time) to acceleration (which is what the question is asking for).

If the divers' speed is constant, then what is their acceleration? What then can you say about the net force acting on them?

3. Oct 6, 2005

can someone tell me if i'm doing this right for part A?

Fnet = 102(9.8) - (.25)(102)(9.8)
Fnet = 999.6 - 249.9
Fnet = 749.7 = ma

so if i'm doing this right..where do i go from here?

4. Oct 6, 2005

### Tom Mattson

Staff Emeritus
So far, so good. Just make sure that you understand that you have elected to take the downward direction as positive. Nothing wrong with that, but you'll find that most physics books choose the opposite convention.

What does the question ask you for?

5. Oct 6, 2005

so i know i'm trying to find the acceleration..so would i plug in the answer i got for the mass?

Fnet = 749.7 = ma
Fnet = 749.7a?? but how can i solve from here if i dont know the Fnet and i'm looking for a?

6. Oct 6, 2005

### cscott

You do have the net force! You even stated it:

So you now know 749.7 = ma or a = 749.7/m

Make sure you understand the definiton of net force.

You set 749.7 = m which isn't correct - think of units.

Last edited: Oct 6, 2005
7. Oct 6, 2005

thanks tom and cscott!! i understand it much better now

8. Oct 6, 2005

### cscott

Not a problem.

9. Oct 6, 2005

### Tom Mattson

Staff Emeritus
Have you taken algebra? You can't just move the "a" from one side of the equals sign to the other like that.