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Homework Help: Max Height

  1. Nov 18, 2007 #1
    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.
  2. jcsd
  3. Nov 18, 2007 #2
    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?
  4. Nov 18, 2007 #3

    Shooting Star

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    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.
  5. Nov 18, 2007 #4
    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?
  6. Nov 18, 2007 #5
    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.
  7. Nov 22, 2007 #6

    Shooting Star

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    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.
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