## max Height

A .050-kg ball is shot at a velocity of 12 m/s by an upward-slanting cannon. At its maximun height, the kinetic energy of thew ball is 24J.

a) Tha max. height is?
b) the angel of the cannon to the horizontal is?

I really dont know where to start on this question. If someone can give me the equation I should used, I can go from there. I am thinking that I have to work backwards using the kinetic energy.
 ok, this is what i figured out so far mgh=1/2mv^2 which m can cancel out gives gh-1/2v^2 which can give me h which is h= v^2/2g which I can solve for is 12^2/ 2*9.87m/s = 7.3m is this right?
 Recognitions: Homework Help If b is the angle in radians of the cannon to the horizontal, then the initial velos ux=12cos b,and uy = 12sin b. The horznt comp ux = vx remains const. At the highest pt, all the kinetic energy is just 0.5*m*vx^2, which gives you vx^2. From that, you can find uy^2, by considering the initial speed. If initial vert velo is known, you can find the max height, and tan b can be found from uy and ux. I have perhaps given too much of detail.

## max Height

ok, i found the max height already, and it was 7.3.

So I think you are saying that using that i can find the angle b , using what equation?