Initial Velocity

  • #1
themadhatter1
140
0

Homework Statement


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

Homework Equations


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

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 traveling at a higher velocity to reach 600m. Do I need to calculate the mass some how? Can someone point me in the right direction?
 

Answers and Replies

  • #2
Phyisab****
586
2
Obviously, the initial velocity depends on the mass of the ball because a cannonball with more mass will need to be traveling at a higher velocity to reach 600m.

Are you sure about that?
 
  • #3
Phyisab****
586
2
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 :smile:.
 
  • #4
themadhatter1
140
0
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 :smile:.

Yeah, your right.

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:
  • #5
Phyisab****
586
2
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
themadhatter1
140
0
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
Phyisab****
586
2
What is the relationship between the kinetic energy immediately after launch and the potential energy at 600m?
 
  • #8
themadhatter1
140
0
they're equal
 
  • #9
Phyisab****
586
2
Exactly just write that and you're almost done.
 

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