1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Max Height
  1. Max height (Replies: 1)

  2. Max height (Replies: 5)

  3. Max Height of Ball (Replies: 2)