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Homework Help: Projectile motion cannon ball angle

  1. Jan 17, 2014 #1
    I need a bit of help with this simple question, it's not difficult at all but I just can't see what I did wrong. I'm overlooking a crucial thing

    A cannon is placed 40m away from a 37.5m high cliff.
    The cannon fires a ball with an angle of 45 degrees to the horizontal.
    The gravitational acceleration is 10 m/s^2.

    The question is: with what mininum initial speed must the ball be fired so that it just hits the top of the cliff?

    I tried to split the velocity in two components, both vx and vy.
    vx = v0x * t (since there is no net force in the x-direction)
    vy = v0y - gt (the only force that acts on the ball is gravity)

    Since the ball is at it's highest point it's velocity in the y direction must be equal to 0. In that case, v0y = gt, so
    t = v0y/g = v0y/10.
    I can substitute this time in the equation for the y- position of the ball and find what the initial speed must be to reach that height
    Sy = v0y*t - 1/2 * g * t^2
    = v0y*t - 5*t^2
    since t = v0y/10, and sy must be equal to 37.5
    37.5 = 5/100 * v0y^2, v0y = sqrt(750) = 27.39
    So v0 = 27.39/sin (45) =38.73 m/s

    However, my textbook gives as answer v0 = 80 m/s. What am I doing wrong?
  2. jcsd
  3. Jan 17, 2014 #2
    First off, I see a with your [itex]v_{0x}[/itex], was that a typo? Easy to notice because the units are off.
  4. Jan 17, 2014 #3


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    What you did looks fine. If you now consider what happens in the x-direction, you should find that you are hitting the wall, so to speak; i.e. the ball hits the wall before it is at it's highest point! Draw a picture and reconsider your starting assumption "Since the ball is at it's highest point it's velocity in the y direction must be equal to 0"
  5. Jan 17, 2014 #4
    Ok, after analyzing the problem slightly more I see one flaw in your argument. I did get 80 m/s too. The flaw: you assume that [itex]v_y=0[/itex] when hitting the target. When, in fact, all the question is asking is to hit the top of the cliff. You could be past the highest point or before the highest point, thus you can't conclude that.

    You are approaching the problem correctly though. Breaking it up into [itex]x[/itex] and [itex]y[/itex] components yields two equations of parabolic motion. You should attempt to find the time with information you already know though (Look at my comment above for a hint, if you correct this equation you can find time.) Then substituting into the other equation will allow you to find [itex]v_{0}[/itex] directly.

    The answer will be 80 m/s exact, too. Also, note that

    [itex]\theta=45[/itex] degrees
  6. Jan 17, 2014 #5
    Ah I now realise that my substition for t should have been 40/v0x since the required initial speed to hit the top of the cliff depends on both vertical and horizontal speed. I now got precisely 80 m/s. Thanks a lot, I feel a bit silly now.
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