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Hard projectile physics question

  1. Apr 19, 2003 #1
    I'm asking if you know the equation to hit a moving target with a projectile.

    the canon, or where the projectile is launched from is 0,0
    the target is at d,0
    the target is moving away at a speed of vx2,0
    Note: the target is always going to be at ground level, 0, and only has a constant x speed.
    the canon it pointing at an angle of a
    the question now, is to solve for the power of the cannon, v.
    y = projectile y value
    vy1 = y velocity of projectile
    basic projectile physics:
    y = vy1*t - (1/2)*g*t^2
    so when y is the height of the target:
    0 = vy1*t - (1/2)*g*t^2 so...
    vy1*t = (1/2)*g*t^2 and...
    vy1 = g*t/2 so that is the y velocity the projectile need to be launched at in order to land in time t.

    since time is distance over rate, the time of collision is t = d/(vx1 - vx2)

    so time of collision with ground need to equal time of collision with target.
    So substitute t = d/(vx1 - vx2) with vy1 = g*t/2
    and you get:
    vy1 = g*d/(2*(vx1 - vx2))
    Note: if vx1=vx2 then vx1-vx2=0 and then the denominator is zero, so they would never collide.
    And we can find the launch speed and angle of the projectile:
    vx1=v*cos(a)
    vy1=v*sin(a)
    v=sqrt(vx1^2 + vy1^2)
    so
    v*sin(a) = g*d/(2*(v*cos(a) - vx2))
    so v = g*d/(2*(v*cos(a) - vx2))/sin(a)

    but there is still a v on the right, could someone help me here in solving for v?
     
    Last edited by a moderator: Apr 19, 2003
  2. jcsd
  3. Apr 19, 2003 #2
    After a couple hours, I figured it out, so I guess I didn't need help afterall. Incase you wanted to know the equation after reading the above post, it is:
    v=(v2/cos(a) + SQRT((-v2/cos(a))^2 + 4*(g*d/sin(2a))))/2
     
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