# Parachute Problem

1. Aug 8, 2008

### zpmodel

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

A parachutist whose total mass is 100kg is falling at 50m/sec when his parachute opens.
His velocity drops to 10 m/sec in a vertical displacement of 40m. What force was exerted by his parachute?

2. Relevant equations:

F = ma
Fg = m(9.8)

This is my first physics course that I am taking and am not very good with the subject as of yet. I in fact have no idea what to do. I've looked at other similar problems on this forums but I could not understand the solution. I am seeking for a learning experience here so that I would be able to solve other problems as well. Please, any help would be greatly appreciated.

2. Aug 8, 2008

### Staff: Mentor

Start by identifying the forces acting on the parachutist. What's his acceleration? Apply Newton's 2nd law.

3. Aug 8, 2008

### zpmodel

Alright. Forces on parachutist, that would be 100kg times the force of gravity, so 980N down? But I think that's his wieght, not his force...

His acceleration is his force divided by his mass, but I am only given the velocity. But I could find the acceleration by finding the slope of the velocity...
The time that it took him to drop to 10m/sec was 4 seconds ( V = d / t )
So by examining it on a graph, the slope is 10 (rise over run, which was 40/4).
So now that I have his acceleration (I think), I could find his force?
F = 100 x 10 = 1000N down?

And then there's the drag force (I think) pushing the parachute back up, which I have no idea how to do.

I don't think I'm on the right track?

4. Aug 8, 2008

### Staff: Mentor

His weight is a force acting on him, but it's only one of the forces acting on him. The other force is the one you are asked to find.

How did you determine that time? You can find the time by considering his average speed during his 40 m fall.

Once you find the time you can find the correct acceleration. (You can also find the acceleration using various kinematic formulas, but any way will do.)

Once you have the acceleration, use Newton's 2nd law to set up an equation and solve for the force of the parachute on the man.

5. Aug 8, 2008

### zpmodel

Ohhhh yeaah:

V2 = V1 + (a)(t)
10 = 50 + (a)(t)

Correct?

But why isn't the time 4 seconds? I thought that the velocity was the distance divided by time; so:

10m/sec = 40m / seconds
seconds = 4

Nevertheless, there's this kinematic equation as well:

V22 = V12 + 2(a)(d)
502 = 102 + 2(a)(40)
a = 30 m/s/s

So, going back to Newton's second law:
F = ma
F = (100)(30)
F = 3000N down

But the question was "What force was exerted by his parachute?". I don't think I know how to calculate that one...

6. Aug 8, 2008

### Staff: Mentor

Sure, but since both a and t are unknown, not immediately useful.

Since the speed is changing, you need to use the average speed to find the time. (10 m/s is the final speed.) What's the average speed?

That works. (But go back and redo it using your original method so you get it correct.) What direction is the acceleration, by the way?

OK, except for the direction. The net force on the man is upward.

Realize that the force you calculated is the net force on the parachutist. Express that net force in terms of the two individual forces acting on him, then you can solve for the unknown force. Pay attention to directions and signs.

7. Aug 8, 2008

### zpmodel

I realised I had V2 and V1 mixed around, it should have been:

V22 = V12 + 2(a)(d)
102 = 502 + 2(a)(40)
a = -30m/s/s

Which makes sense now because he is accelerating downwards (negative).

F = ma
F = (100)(-30)
F = -3000N upwards

Hmm, so the net force is -3000N. So,
Fnet = F1 + F2
-3000 = F1 + F2

So if I could now find the force of the man, F1, I could find the other force. The man has a gravatational force, (100kg)(-9.8) = -980N. So...

-3000 = F1 + F2
-3000 = -980 + F2
F2 = -2020N

But something still confuses me. Why is the net force on the man upward? Isn't his direction downward? Isn't his position going in the negative direction over time?

8. Aug 8, 2008

### Staff: Mentor

The acceleration is upwards. Let's be careful with signs and call up positive and down negative. Note that the distance traveled is 40 m down, thus should be -40. So:
(-10)^2 = (-50)^2 + 2(a)(-40), thus a = + 30 m/s/s.
Using our sign convention, that net force should be +3000 N (upwards). From Newton's 2nd law we know that the net force and acceleration must be in the same direction.
That should be: +3000 = F1 + F2.
Good, the gravitational force does point downward and thus is negative.
Correct this: +3000 = -980 + F2.
While he is moving in the negative direction, he's accelerating in the positive direction. He starts out with a velocity of -50 m/s and ends up with a velocity of -10 m/s. That's a change of (-10) - (-50) = +40 m/s, a positive change. (If he was accelerating downward, he'd be speeding up as he fell--not good for a parachutist!)

9. Aug 8, 2008

### zpmodel

Problem solved; Sweet! What an awesome forums. I just started in the science field so I'll be probably be around for the next few years.

Thanks a lot for all your help!