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Projectile motion, 30 and 60 degrees on a level plain

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    prove that two bullets fired at 30 and 60 degrees will travel the same distance, and the 60 degree shot will travel 3 times as high as the 30 degree.

    2. Relevant equations
    S(displacement) Vo(initial velocity) Vf(final Velovity) t(time) a(acceleration due to gravity)

    S=Vot + .5at^2
    S=Vft - .5at^2
    S= (Vft - Vot)/2
    Vf^2= Vo^2 + 2aS
    Vf= Vo +at

    3. The attempt at a solution

    ive tried numerous ways, but i cant seem to get past point of the final velocity at the peak being zero to solve for time in the last equation...and i cant figure out what to do from there
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 18, 2007 #2
    The motion is in two directions; the x- and y-directions. At the peak the speed in the y-direction is zero, but there is still speed in the x-direction. You need to analyse this problem in two-dimensions.
     
  4. Oct 18, 2007 #3
    Hi

    To prove that the 60° shot will travel 3 times as high as the 30°:

    You can use the conservation of energy: Kinetic Engergy at the time of the shot equals the potential energy at the peak.

    [tex]E_{kin}=E_{pot}[/tex]

    For [tex]E_{kin}[/tex] you have to take the angle into account:
    [tex]E_{kin}=\frac{1}{2}*m*v_{0}^{2}*sin(\alpha)^{2}[/tex]

    Solve for h (=max height) and compare the values for 30 and 60 degrees.

    I hope this helps...
     
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