Finding the Initial Velocity and Angle of a Cannon Ball Launched from a Cliff

[safat]

Homework Statement

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.

Homework Equations

R=vt+1/2*a*t²
finalv=v+a*t

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..
any hint please?

 

[Doc Al]

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.

 

[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.

 

[Doc Al]

Yep
I got v=17.7ms^-1

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

 

[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.

 

[Doc Al]

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.

Yes.

 

[safat]

great.
thanks for your help!