# Nonconservative forces/ and conservation energy problem

1. Oct 2, 2007

### logglypop

An 80kg sky diver jump out of a balloon at an altitude of 1000m and opens the parachute at an altitude of 200m.a) assuming that the total retarding force on the diver is constant at 50N with the parachute closed and constant at 3600 with the parachute open, what is the speed of the diver when he lands on the ground?b) do you think the sky diver will get hurt? c) at what height should the parachute be opened so that the final speed of the sky diver when he hits the ground is 5m/s

plz help me out
sincerly

2. Oct 2, 2007

### Kurdt

Staff Emeritus
Welcome to PF. Unfortunately the guidelines of the forums do not allow us to give solutions when you have shown no attempt at the question yourself. The guidelines are here: https://www.physicsforums.com/showthread.php?t=5374

What equations do you think you need to solve this problem?

3. Oct 2, 2007

### logglypop

1/2mv^2 and mgh
i try problem 2 time,
my answer is different from book but they doesn't show me how to there

4. Oct 2, 2007

### Kurdt

Staff Emeritus
5. Oct 2, 2007

### logglypop

yes i seen those equation this is wat i do for a). i use velocity and displacement to find the Velocity as he hit the ground
v^2=V^2+2aX
v^2=0+2(9.8)(1000-200)
i got V= 125

did i do anything wrong?

6. Oct 2, 2007

### Kurdt

Staff Emeritus
For part a you are going to have to do it in two stages. The first stage is where the parachutist has not opened his parachute and the second is when he has opened his parachute. Remember that he hasa constant retarding force in both cases which means he will not accelerate at the full 9.81 ms-2 due to gravity.

7. Oct 2, 2007

### logglypop

F=ma
a=50/80=.625m/s^2
v^2=V^2+2aX v^2=0+2(.625)800=31.6m/s this is the final velocity of his 1st part where he open the parachute. correct me if i wrong

8. Oct 2, 2007

### Kurdt

Staff Emeritus
The acceleration he experiences will be 9.81 - 0.625 ms-2 ok?

9. Oct 2, 2007

### logglypop

The acceleration he experiences will be 9.81 - 0.625 ms-2 ok?
why is that?

i did like u say
v^2=V^2+2aX v^2=0+2(9.81-.625)800
v=121.2m/s

10. Oct 2, 2007

### Kurdt

Staff Emeritus
Now the second part is when he opens the parachute. Work out what the resultant acceleration will be then. The reason the acceleration is not simply that due to gravity is the fact that he is experiencing a retarding force. That is a force that slows him down, and thus the effective acceleration is slightly less.

11. Oct 2, 2007

### logglypop

v^2=V^2+2aX v^2 a=3600/80= 45 9.8-45= -35.2m/s^2
v^2=121+2(35.3)200
v=119m/s

according to the book answer they have 24.5m/s
what did i do wrong?

12. Oct 2, 2007

### Kurdt

Staff Emeritus
You've used a positive value for the acceleration when it should be negative because the diver is slowing down now.

13. Oct 2, 2007

### logglypop

if i use negative for acceleration then i got the square root of negative number for velocity

14. Oct 2, 2007

### Kurdt

Staff Emeritus
No you don't because (121.2)2 is bigger than 400 x (-35.2).

15. Oct 2, 2007

### logglypop

im sorry i forgot to square the initial velocity. I feel bad now

16. Oct 2, 2007

### Kurdt

Staff Emeritus
Its fine I've done it myself.

17. Oct 2, 2007

### logglypop

c) v^2=V^2+2aX
5^2=v^2+2(-35.2)x

i need to find initial Velocity to calculate x, im stuck again
give me hint plz

18. Oct 2, 2007

### Kurdt

Staff Emeritus
Well you know the initial velocity will be given by $v^2 = u^2+2(9.185)x_1$

You also know $x_1 = 1000 - h$ and $x_2 = h-0$ and thus you can substitute into the equation you have given to get an equation of only one variable which you can solve.