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Projectile Olympiad Question

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data

    A cannonball is launched with initial velocity of magnitude v0 over a horizontal surface. At what minimum angle θmin above the horizontal should the cannonball be launched so that it rises to a height H which is larger than the horizontal distance R that it will travel when it returns to the ground?

    (A) θmin = 76◦
    (B) θmin = 72◦
    (C) θmin = 60◦
    (D) θmin = 45◦
    (E) There is no such angle, as R > H for all range problems.


    2. Relevant equations

    d = (vi+vf)/2)*t

    3. The attempt at a solution

    H = (1/2)(v0sinθ)(t) and R = (v0cosθ)(t)

    Thus, if H = R, then (1/2)(v0sinθ)(t) = (v0cosθ)(t)
    =>tanθ = 2, so θ = 63.4°. I'm probably making a really obvious mistake here, but I'm not seeing it. Any help would be appreciated.
     
  2. jcsd
  3. Jan 9, 2013 #2

    TSny

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    Do you need to include the effect of gravity somewhere? [EDIT: Nevermind, you are using the equation d = (vi + vf)*t/2 which doesn't require knowing the acceleration.]
     
    Last edited: Jan 9, 2013
  4. Jan 9, 2013 #3

    TSny

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    I think I now see what you're doing. Does the t in the H equation represent the same time as the t in the R equation?
     
  5. Jan 9, 2013 #4
    ohhhh i see now. the t in the H equation is the time to get to maximum height, which is half of the t in the R equation. so it would really be (1/2)(v0sinθ)(t/2) = (v0cosθ)(t) which gives tanθ = 4 so θ = 75.9° (A). Thanks so much!!
     
  6. Jan 9, 2013 #5

    TSny

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    Good job.
     
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