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Math Help - Projectile Motion

  1. #1
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    Projectile Motion

    We have a cannon that shoots out a ball, and we have to make it hit a target on the floor.

    We have to aim our cannon 25.0 degrees, and it is placed on a ledge 1.51m off the floor. The ball has an initial velocity of 4.534m/s. I think we have to find the x and y component vectors, but I already tried solving it and launching it and we overshot it (our guess was 1.62meters).
    Last edited by vesperka; November 26th 2007 at 11:11 AM.
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  2. #2
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    For me, first I resolved the two component vectors. The hypotenuse is 4.534, so using SOH CAH TOA I found the Y component to be 1.916m/s and the X to be 4.109m/s. Then, I used the distance formula

    Dy = VoT + 1/2(g)(t^2)

    and solved for T to get

    1.51 = (1.916)(t) + 1/2(9.80)(x^2)

    Then T = .393

    Then to solve for Dx I used

    Dx = VoT

    the acceleration part of the equation is eliminated because their is constant uniform velocity on the X component vector.

    Dx = (4.109)(.393)
    Dx = 1.62m

    This answer is wrong though
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  3. #3
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    Quote Originally Posted by vesperka View Post
    We have a cannon that shoots out a ball, and we have to make it hit a target on the floor.

    We have to aim our cannon 25.0 degrees, and it is placed on a ledge 1.51m off the floor. The ball has an initial velocity of 4.534m/s. I think we have to find the x and y component vectors, but I already tried solving it and launching it and we overshot it (our guess was 1.62meters).
    The reason it isn't working for you is you are setting it up wrong.

    I have a coordinate system with an origin at the point directly below (1.51 m below) where the cannon ball is fired. I have a +x direction to the right (the direction I'm firing the ball in my diagram) and a +y direction upward.

    Here's what we know:
    t = ? <-- This is the time when the ball hits the floor.
    x_0 = 0~m~~~~x = ? y_0 = 0~m~~~~y = 0~m
    v_{0x} = v_0~cos(25.0)~m/s~~~~v_x = v_{0x}~~v_{0y} = v_0~sin(25.0)~~~~v_y = ?
    a_x = 0~m/s^2~~~~a_y = -9.8~m/s^2

    We want x, so
    x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2

    x = v_{0x}t
    which we can't solve for since we don't know t.

    So
    y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2

    0 = 1.51 + v_{0y}t - 4.9t^2

    We can solve this for t:
    t = \frac{-1.91615 \pm \sqrt{1.91615^2 - 4 \cdot -4.9 \cdot 1.51}}{2 \cdot -4.9}

    t = 0.784078~s
    disregarding the negative solution.

    So we have:
    x = v_{0x}t = 3.22193~m

    Always organize your data before you start working the equations!

    -Dan
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