# Homework Help: Horizontal cannon on a cliff

1. Oct 9, 2012

### safat

1. The problem statement, all variables and given/known data

Classic cannon ball launched with an initial horizontal velocity from an elevated position.
Initial hvelocity: v
Final velocity (when hits the ground): 4v
High of the cliff: 16m
Find the initial velocity and the angle at which the ball strikes the ground.

2. Relevant equations

R=vt+1/2*a*t2
finalv=v+a*t

3. The attempt at a solution
First, I find the vertical/horizontal info
initial horizontal velocity:v initial vertical velocity:0
horizontal acc:0 vertical acc:-9.82
horizontal velocity at instant t:v vertical velocity at instant t:gt
horizontal displ: v*t vertical displ:16??

I find t=√2s/a using the vertical info, but then I get stuck..

Last edited: Oct 9, 2012
2. Oct 9, 2012

### Staff: Mentor

Find the vertical component of the final velocity. (First in terms of v.)

Then you might want to use another kinematic formula to relate speed and distance for accelerated motion.

3. Oct 9, 2012

### safat

Yep
I got v=17.7ms-1

I am really tempted to do 17.7/4 => u=4.4ms-1, but I guess it's far too easy and sounds wrong.
I have also noticed that if I put v and h as vectors nose to tail, I have 4v as a resultant vector. I am kind of sure this is the way for finding the angle of impact.

4. Oct 9, 2012

### Staff: Mentor

That's the vertical component of the final velocity in m/s, but that's not what the problem is calling 'v'. Express this speed in terms of 'v'.

Hint: Pythagoras

5. Oct 9, 2012

### safat

Ok, sorry for the poor quality of the sketch, but I want to be sure I understand this problem.
Am I right by doing so?

Thus I can find the angle between 4v and v, as the angle of impact.

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6. Oct 9, 2012

Yes.

7. Oct 9, 2012

great.