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Height of a projectile at two points?

  1. Jan 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired with a velocity v such that it passes through two points both at a distance h above the horizontal. Show that if the gun is adjusted for maximum range, the separation of the points is

    d = (v/g)(√v2-4gh)

    2. Relevant equations
    y = -gt2/2 + vsinθt
    x = vcosθt

    for max range R = v2/g
    θ = π/4

    3. The attempt at a solution
    substituting theta into the two equations gives me
    h = -gt2/2 + vt/√2

    d = (v/√2)t

    I replaced y with h to find the two times at which the ball is at h (both going up and down), then I used those times I found in the d equation to find the distance at both points at both times, and then took the difference between "d1" and "d2" where d2 is the point after max height and d1 is the point before. Is this not the right method? I get that t = h and t = stuff from the h equation... and t = h is obviously not true. Is that the issue?
     
  2. jcsd
  3. Jan 11, 2017 #2

    haruspex

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    Looks right.
    Then you are making an algebraic error. You can use dimensional analysis to see where it goes wrong. If you cannot spot it, please post all your working.
     
  4. Jan 11, 2017 #3
    h = -gt2/2 + vt/√2 to begin. I pull a t out and it becomes h = t(-gt/2+v/√2) so h = t and -gt/2+v/√2 = h

    :/
     
  5. Jan 11, 2017 #4

    haruspex

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    By what logic?
     
  6. Jan 11, 2017 #5
    That looks suspiciously like what you would do if you were trying to find the zeros of an equation. i.e. t(-gt/2+v/√2) = 0 so t = 0 or
    (-gt/2+v/√2) =0. That isn't at all correct for h.
     
  7. Jan 11, 2017 #6
    Finished. Thank you.
     
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