1. Nov 18, 2007

### splac6996

1. The problem statement, all variables and given/known data
A projectile is shot straight up from the earth's surface at a speed of 1.40×10^4 . How high does it go?

2. Relevant equations
potential energy is given by (GM(1)M(2))/R

3. The attempt at a solution
I attempted to solve this using energy conservation equation and solving for r which I thought would be the radius of the earth plus the distance traveled by the projectile. By subtracting the radius I thought I would get the distance traveled but my answer is wrong can someone help please

2. Nov 18, 2007

### dynamicsolo

What units is that speed in?

It might be of help to a reader who is going to assist you if you would show the actual calculation you did.

3. Nov 18, 2007

### Mindscrape

There are a number of things you could have done wrong. For example, did you remember to use a negative sign in your potential? Do you have an initial potential, where did you define your zero in potential energy to be? Specifically, what was the equation you used?

4. Nov 19, 2007

### splac6996

the speed is defined in km/hr and for the formula that i used I did use a negative sign for my potential energy and I used my zero point to be the point were the projectile is shot from.

5. Nov 19, 2007

### katchum

Why not just use m.g.h=1/2mv^2?

And potential energy is in Joule not in Newton.

6. Nov 19, 2007

### Kurdt

Staff Emeritus
mgh is only applicable near the Earth's surface. For this projectile it will be travelling significantly far away from the Earth's surface to make mgh non-applicable.

7. Nov 19, 2007

### splac6996

Well my understanding is that if the distance is smaller than than the radius then I could use mgh=1/2mv^2 but in this case since the height is going to be largerthan the radius of the earth so my answer would be wrong.

8. Nov 19, 2007

### Oerg

use v^2=u^2+2as,

your deceleration is not constant in the equation. The deceleration is given as the average deceleration the object undergo in that motion. to find the average deceleration,

integrate the below equation with respect to distance, with the radius of the earth as the lower limit and the radius+height of the earth as the upper limit.

then divide by the height to find the average deceleration.

a=Gm/r^2

EDIT: Well, you will need to have the radius of the earth to solve the equation. Even after simplifying expressions of the radius of the earth into g.

Last edited: Nov 19, 2007
9. Nov 19, 2007

### Sourabh N

@splac6996: Well, If you take zero of your PE at surface of the Earth, your formula for PE at R (in #1 post) is not correct.

Last edited: Nov 19, 2007
10. Nov 19, 2007

### splac6996

this is what i think
-GM(earth)m(projectile)/H(radius of earth + distance traveled)=(1/2)m(projectile)v^2
doing the algebra and simplifying i have
H=-2GM/v^2
is that correct

11. Nov 19, 2007

### Sourabh N

Nope. You can see, if you put distance traveled = 0 your
RHS = 0 but LHS is not = 0.

12. Nov 19, 2007

### Oerg

the expression for the potential -GM(earth)m(projectile)/H(radius of earth + distance traveled) gives you the potential energy at the highest point. You will still need to minus the potential at the earth's radius.

13. Nov 19, 2007

### HallsofIvy

Staff Emeritus
With gravitational force $F= -GMm/r^2$, the potential energy is $GMm/r$ but that has 0 point at infinity, not at the surface of the earth.

14. Nov 19, 2007

### Sourabh N

Can you show the exact figures you got?