# Initial Velocity

1. Jan 18, 2010

1. The problem statement, all variables and given/known data
A cannon shoots straight up into the air a cannonball which reaches a maximum height of 600m. What was the initial velocity of the cannonball when it was fired from the cannon. (Note: this problem contains all the information necessary to be solved.)

use g=10m/s^2

2. Relevant equations
P.E= mgh
K.E= (1/2)(k)(x^2)

3. The attempt at a solution

P.E=m(10m/s^2)(600m)

I have no clue as to how you would solve for the initial velocity without the mass of the cannon ball. Obviously, the initial velocity depends on the mass of the ball because a cannonball with more mass will need to be travelling at a higher velocity to reach 600m. Do I need to calculate the mass some how? Can someone point me in the right direction?

2. Jan 18, 2010

### Phyisab****

3. Jan 18, 2010

### Phyisab****

Also your equation for kinetic energy is for the potential energy of a spring with spring constant k displaced x from equilibrium, I don't think that's the equation you wanted .

4. Jan 18, 2010

I meant to put K.E.= (1/2)(m)(v^2)

I still don't understand what I'm suppose to do. Both equations require mass to solve for the potential or kinetic energy. Does mass matter in this problem because I don't think I can solve for the mass of the cannonball?

Last edited: Jan 18, 2010
5. Jan 18, 2010

### Phyisab****

Nope mass does not matter, remember the story of galileo dropping ball from the tower of pisa? The equation you just wrote is for the kinetic energy not potential energy. See what happens when you put them together.

6. Jan 18, 2010

I'm sorry but I don't know how you would combine these equations. I have tried to combine them but I come up with an answer that you wouldn't be able to solve for the v^2 and dosent seam logical.

K.E=(1/2)(m)(v^2)
P.E.= mgh

P.E/(G)(H)=m

K.E=(1/2)[P.E/(G)(H)](v^2)
K.E=(1/2)[P.E/(10)(600)](v^2)
K.E=(1/2)[P.E/6000](v^2)
K.E/(1/2)[P.E/6000]=v^2

Somehow I don't think the square root of the left side is the answer. What am I doing wrong?

7. Jan 18, 2010

### Phyisab****

What is the relationship between the kinetic energy immediately after launch and the potential energy at 600m?

8. Jan 18, 2010