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Proof: maximum range of an object launched off a cliff

  1. May 2, 2012 #1
    I've been working on this for a while, with little luck. We are told that an object is on a cliff of height H, and is launched off with a velocity v at an angle of theta with the vertical. gravitational acceleration is g. To get the maximum range, it must be a certain angle theta, given by the formula:
    cos^2(θ)=(v^2)/(2v^2+2gh). Prove that this is correct.



    v*sin=range/t, -h=v*cos*T-.5gT^2, getting a quadtratic theorem, and deriving. No luck, I get ugly giant equations rather than that small solution.
     
  2. jcsd
  3. May 3, 2012 #2

    tms

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    First find out how much time is needed. Do that by looking at the vertical component. The total time is the time it takes to go up to its maximum height, plus the time it takes to fall back to the original height, plus the time it takes to reach the ground.
     
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