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Projectile Motion help

  1. Oct 6, 2013 #1

    d21

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    1. The problem statement, all variables and given/known data
    A ball is thrown off of a 15m cliff at an angle of 30 degrees above the horizontal and strike the ground 10.4 seconds later. Determine the initial velocity of the ball, the maximum height of the ball, and impact (final) velocity of the ball.


    2. Relevant equations
    vo=vf+at
    y=.5at+vot
    x=vot


    3. The attempt at a solution
    I'm really out of my depth on this problem. I really just have no clue where to begin, since I'm used to being given the initial velocity of the ball. I had an idea to try to divide up the time of the problem into two sectors, where the ball is in its parabolic motion in one section and the other where it comes back to the same level that it was launched at and continues to the ground. the time would be √2h/g correct? But can I use this even though there's a y component to the velocity? Is any of this even correct? How do I do this problem?

    Or can I just say y=.5gt^2+vot
    -15=.5(-10)(10.4)^2+vo(10.4)?
    Vo=53.44?

    Also isn't this just the vertical component of Vo? So shouldn't the full Vo be
    Voy=VoSin30
    53.44=VoSin30
    Vo=106.88?
     
    Last edited: Oct 6, 2013
  2. jcsd
  3. Oct 7, 2013 #2
    well, you do have to work in sections. when the ball comes back to the same level on the other side, it has the same magnitude of vertical velocity (the direction is opposite) it had at launch time. from here on, you know the height to the bottom, calculate final vertical velocity.
    Use the equation
    V^2 - Vo^2 =2as

    it should become simple from here on.
     
  4. Oct 7, 2013 #3

    Simon Bridge

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    That would be if you were using the standard ballistics equations.
    Alternately - draw a v-t diagram for horizontal and vertical components of the motion.
    Or just applying the kinematic equations, as d21 has done.

    Welcome to PF;
    Proceed as you normally would, just leave the initial velocity as a variable.
    You'll end up with a system of simultaneous equations which you can solve for the initial speed.

    As NahilSH said, you can finess the process by realising that the speed is the same at the initial height but coming down.
     
  5. Oct 7, 2013 #4
    yeah, true...:redface: but it does help to work in sections to simplify the process.
     
  6. Oct 7, 2013 #5

    Simon Bridge

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    It can help to break the process into stages, and solve each stage in turn. Yep.
     
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