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Projectile Range and Height

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    At 0.5 of its maximum height, the speed of a projectile is 0.75 of its initial speed. What was its launch angle?

    2. Relevant equations
    Not sure here, but:
    v^2 - vo^2 = 2as [maximum height]
    Ay = Asin(theta) [y-component]

    3. The attempt at a solution
    I didn't get very far, but:
    vo = vo
    v = 0.75 vo

    0.75vo^2 - vo^2 = (2)(-9.8)s
    -0.25vo^2 = -19.6s
    vo^2 = 784s

    I'm stuck from here on, should I try to find the y-component for this?
    [Also I'm an almost total beginner at Physics, excuse me if I'm totally wrong].

    Any help would be greatly appreciated. Thanks everyone!
     
  2. jcsd
  3. Oct 4, 2015 #2
    I thought about it a little more:

    x-component : [v0^2/g]sin(theta)
    y-component : vo sin(theta)

    (theta) = tan^-1 [vo sin(theta)] / [vo^2/9.8 sin (theta)]
    (theta) = tan^-1 (9.8/vo)

    Is this headed in the right direction? If so, what would I substitute in for vo? If not, any tips?
    Thanks!
     
  4. Oct 4, 2015 #3
    Oh wait I think it would be best to use energy conservation.

    0.5 mv^2 = 0.5m(0.75v)62 + 0.5mgh
    and then
    0.5 mv^2 = 0.5 m(v cos theta)^2 + 7/16(mv^2)
    and simplify to get theta = 69 degrees

    Is this correct?
     
  5. Oct 5, 2015 #4
    I got the same value for θ0 (but I don't really understand your way to find it). So I think the answer is correct, but the calculation is confusing.
     
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