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Projectile Motion launch angles

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A projectile is fired in such a way that its maximum height is a factor of 4 times its
    horizontal range. Find the launch angle in degrees

    2. Relevant equations
    [itex]h_{max}=\frac{V^{2}_{0}sin^{2}\theta}{2g}
    Range=\frac{V^{2}_{0}sin2\theta}{g}[/itex]

    3. The attempt at a solution

    I set h=4R giving
    [itex]sin^{2}(\theta)=8sin(2\theta); [/itex]
    [itex]sin^{2}(\theta)=16sin(\theta)cos(\theta); [/itex]
    [itex]sin(\theta)=16cos(\theta); [/itex]
    [itex]tan(\theta)=16 [/itex]
    [itex]\theta=86^{o} [/itex]
    but the answer is apparently 38.66 degrees.

    So I assumed that I had misinterpreted the question, and in fact R=4h, however doing this I got an angle of 45 degrees.
    I'm not sure where I've gone wrong
     
  2. jcsd
  3. Jan 18, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi Jon.G. Welcome to Physics Forums.

    Given the problem statement as it stands your interpretation and calculations look fine to me (using a bit more precision, θ = 86.42°). Are you quoting the problem statement exactly? It could be that they changed the problem statement slightly, to make it a "new" question, but failed to change the answer key.
     
  4. Jan 19, 2014 #3
    I emailed my tutor and that is exactly what happened :)
     
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