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!

Finding the max height of a ball launched as a projectile using work-energy

  1. Mar 19, 2007 #1
    A ball is launched as a projectile with initial speed v at an angle theta above the horizontal. Using conservation of energy, find the maximum height h_max of the ball's flight.
    Express your answer in terms of v, g, and theta.

    My energy equation is as follows:

    0.5m(v^2)cos(theta) +0.5m(v^2)sin(theta) = 0.5m(v^2)cos(theta) + mgh_max

    v^2cos(theta) + v^2sin(theta) = 2gh_max + v^2cos(theta)
    h_max = (v^2sin(theta))/2g

    This isn't right, somehow. I know that the only velocity at max height is in the x-direction, so max height should be a sine function, right? Did I forget to include something in my energy equation?
  2. jcsd
  3. Mar 19, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks like you forgot to square the trig functions. At the top of the flight, for example, the KE is
    [tex]KE = 1/2m(vcos\theta)^2[/tex]. Also, in the initial case, you don't have to break v into its x and y components. You can just use 1/2mv^2 and get the same result.
  4. Mar 19, 2007 #3
    thanks! i got

    h_max = (v^2sin^2(theta))/2g

    You actually cleared up a big confusion for me in general. I always was using v^2trig(theta) in lots of problems and never understood what was going wrong.
  5. Mar 19, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, that is correct, and better than my suggestion to not break up v_initial into its components. In which case you would get, not to confuse you,
    [tex] 1/2mv^2 = mgh_{max} + 1/2m(vcos\theta)^2 [/tex]
    [tex] v^2 = 2gh_{max} + (vcos\theta)^2 [/tex]
    [tex] h_{max} = (v^2 - (vcos\theta)^2 )/2g [/tex]
    [tex] h_{max} = (v^2(1 - (cos\theta)^2)/2g [/tex]
    [tex] h_{max} = (v^2sin^2\theta)/2g [/tex]
    which is a bit more tedious than your approach.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?