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Prove that a projectile launched on level ground

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove that a projectile launched on level ground reaches maximum height midway along its trajectory.

    3. The attempt at a solution

    Capture.JPG

    Is the attached solution the correct way to prove?

    I tried another method which was:
    t = 0 or t = 2vyi/g

    [y(2vyi/g) = y(0) + vi.t -0.5gt^2]/2 [itex]\equiv y(vyi/g) = y(0) + vyi.t -0.5gt^2 = vyi^2/g[/itex]
    then solve but it couldn't be solve.
    This was so that I could prove that the total displacement in t_apex = vyi/g is equivalent to half the displacement achieved in y(2vyi/g) = y(0) + vi.t -0.5gt^2
     
  2. jcsd
  3. Jan 25, 2014 #2

    phinds

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    (1) Just to be absolutely clear, rather than clear by assumption, the problem statement should say that this is a ballistic trajectory (it always is in such problems, but still ... )

    (2) I interpret the question to mean that you are to show that the apex occurs when half of the total horizontal distance has been covered. I think this would be a more normal interpretation that the other two possible interpretations which are "when half of the total time has elapsed" and "when BOTH half of the time AND half of the distance have elapsed" (which happens to be the case but to my mind wasn't really the question).
     
  4. Jan 25, 2014 #3
    For half the horizontal distance to be covered, half the time from x=0 to x = x must have elasped. I have found this time, symbolically. I subbed this time into the y-displacement component. Would this, thus, demonstrated the proof?
     
  5. Jan 25, 2014 #4
    I hope you know the formula for the range of a symmetric projectile. You also know that time of ascent equals time of descent. Hence when y=H=Maximum height, x=Horizontal velocity * (total time)/2 = ....
     
  6. Jan 25, 2014 #5
    I know how to derive those equation.
    What I did in the attachment was to find half the time required for the trajectory, then sub that time into the y-displacement equation. That would give me the displacement of the y-component at t = 1/2.
    What I'm asking is if this alone is a sufficient condition for proving the demands of the question.
     
    Last edited: Jan 25, 2014
  7. Jan 25, 2014 #6
    Well by your method you concluded in a long winded way that time of ascent = total time/2 = vy/g..

    But then how you will prove that at y=maximum height, x= Range/2 for a symmetric projectile. You will need an equation for horizontal.
     
  8. Jan 25, 2014 #7
    Am I required to use the x-displacement equation? If so, x = vi cos Θ .t
    t at y_apex = vyi/g

    x = vi cos Θ (vyi /g)
     
  9. Jan 25, 2014 #8

    phinds

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    Personally, to show that half the distance is covered, I would find the total horizontal distance H and the horizontal distance to the apex A and show that A = H/2. That seems to be the most unambiguous way to do it.
     
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