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Math Help - Projectile Motion

  1. #1
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    Projectile Motion

    I have been stuck on this for a while....

    A projectile's launch speed is five times it speed at maximum height. Find the launch angle.

    Anybody have any clue how to do this?
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  2. #2
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    Quote Originally Posted by Elite_Guard89 View Post
    I have been stuck on this for a while....

    A projectile's launch speed is five times it speed at maximum height. Find the launch angle.

    Anybody have any clue how to do this?
    Speed at maximum height = U \cos \theta (since the vertical component of the velocity is equal to zero at maximum height).

    Therefore U = 5U \cos \theta. Solve for \theta.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Speed at maximum height = U \cos \theta (since the vertical component of the velocity is equal to zero at maximum height).

    Therefore U = 5U \cos \theta. Solve for \theta.
    i am now confused wat?
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  4. #4
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    \cos(\theta_{max}) = 1 (at maximum height), since this is the parabolic path of a projectile. So you have an unknwn velocity at maximum height, v\cos(\theta_{max}) = v, an, unknown initial velocity, u=5v\cos(\theta), and an unknown angle \theta, which you need to find. I'd say there isn't enough information to find it. if you are assuming gravity of -9.8ms^{-2}, and have the final velocity, the launch and impact heights, then you could plug in the values for x and y, and solve simultaneous equations using the equations of motion.
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