# Cannon hits cliff

1. Feb 2, 2008

### Yahaira.Reyes

A connon, located 60 m from the base of a vertical 25.0m tall clift, shoots a 15 kg shell at 43.0 degrees above the horizontal toward the cliff.

A) what must the minimum muzzle velocity be for the shell to clear the top of the cliff?
B) The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (A), how far does the shell past the edge of the cliff?

x=x0 + V0t
Vy= v0y-gt
y=y0+v0yt-1/2gt^2
vf^2=vi^2-2ax

I am a little throw off by the kg of the shell. I do not know how to start the prob.

2. Feb 2, 2008

### HallsofIvy

As is always the case with "gravity" problems, the mass of the shell is irrelevant. Ignore it. Use precisely the equations you have with one addition: If V0 is the initial muzzle velocity, then the x-component is VO cos(43) and the y-component is VO sin(43).

3. Feb 2, 2008

### Yahaira.Reyes

wouldn't I need to calculate the time ? All of the equations require time.

4. Feb 2, 2008

### Littlepig

no, you can put time in function of x, and then substitute in equation of y, getting y in funtion of x.

that gives the y coordinate in respect with x. with x=60 and y=25,you know that there will be a velocity V0 needed.(this after you've done the cos/sin thing HallsofIvy told and well.)