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2D Projectile Problem

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A cart is moving horizontally along a straight line with a constant speed of 30m/s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80m. At what speed and at what angle to the horizontal must the projectile be fired?


    2. Relevant equations
    The kinematic equations.
    The problem ignores air resistance and friction.


    3. The attempt at a solution
    The first thing I said in my answer to this problem was, "since the cart is moving at a constant velocity, for a projectile to land back on the cart it would have to be fired at 90 degrees to the horizontal". After stating this I could easily solve for the velocity.

    My professor however tells me that this is not true and that the angle will be less than ninety degrees.
     
  2. jcsd
  3. Oct 24, 2009 #2

    Delphi51

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    Homework Helper

    Welcome to PF!

    I think you can write up a proof of your conjecture easily.
    The equation of motion for the cart is x = 30t.
    The equations for the projectile are x = 30t and y = Vi*t - ½gt²
    (put in your number for Vi). Figure out the time when y = 0 and show that both horizontal positions equal 80.

    If you ask the prof to look over your work without being too pushy about it, you'll likely win him over. Demonstrating with the high school apparatus that shows this effect would likely be too pushy.
     
  4. Oct 24, 2009 #3
    The direction of the velocity relative to the cart is indeed straight up. Apparently your professor wants the direction of the velocity relative to the ground. (It's not how I would interpret it).

    You already computed the vertical velocity relative to the cart and the ground. The horizontal velocity is the speed of the cart.
     
  5. Oct 24, 2009 #4

    Delphi51

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    Homework Helper

    Thanks, willem2! I missed that entirely!
    The angle is 90 degrees from the cart's point of view, but from the point of view of the ground, there will of course be a horizontal component to the velocity of the projectile.
     
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