Max Height and Angle of Cannon in 12 m/s Ball Shot with 24J Kinetic Energy

[NewJersey]

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 don't 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.

 

[NewJersey]

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?

 

[Shooting Star]

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.

 

[NewJersey]

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?

 

[NewJersey]

Ok ,
what about this equation
24J= 12cos(degree)

can i set it up like this to find the answer , if so is the answer 66degrees.

 

[Shooting Star]

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?

That's not correct. (Check the data you have given -- doesn't make sense.)

Did you understand what I had said in my earlier post? If not, tell me which portion you didn't understand. I think you should thoroughly revise projectile motion once before attempting problems.