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Peaked Roof

  1. Aug 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A peaked roof is symmetrical and subtends a right angle, as shown. Standing at a height of distance h below the peak, with what initial speed must a ball be thrown so that it just clears the peak and hits the other side of the roof at the same height?

    Diagram: http://imgur.com/bi1efMm

    2. Relevant equations
    [itex]x=vcos(\theta)t[/itex]
    [itex]y=vsin(\theta)t-\frac{1}{2}gt^2[/itex]
    [itex]\frac{1}{2}mv^2=mgh[/itex]

    3. The attempt at a solution
    This is supposed to be a kinematics problem, but I wasn't too sure how to do it that way so I used energy principles. Using the above equation, I got that the velocity should be [itex]\sqrt{2gh}[/itex] by simply rearranging the variables, however, the answer key says that it's [itex]\sqrt{5/2gh}[/itex].
     
  2. jcsd
  3. Aug 24, 2015 #2

    Nathanael

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    That equation you used is wrong. It assumes the speed is zero when the height is h.

    Give it another shot, then explain your thinking.
     
  4. Aug 24, 2015 #3
    Does it have something to do with how the vertical speed is 0 but it is still moving horizontally? In that case the vertical speed at the bottom would be [itex]\sqrt(2gh)[/itex], and the horizontal would be (by working backwards from the answer) [itex]\sqrt(1/2gh)[/itex].
     
  5. Aug 25, 2015 #4

    haruspex

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    Yes.
    That would have dimension 1/speed, so cannot be right.
    Knowing the vertical launch speed, how long does it take to reach the top? How far has it moved horizontally in that time?
     
  6. Aug 25, 2015 #5
    Sorry for bad formatting, I meant [itex]\sqrt(1/2) * \sqrt(gh)[/itex].

    I decided to do the problem a different way, I used the formula for max height and max range, set max height to half of max range, and solved for the angle. From there I was able to get the max height, which indeed was [itex]\sqrt(5/2)\sqrt(gh)[/itex]. Thanks for your help.
     
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