Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    v=sqrt(vx1^2 + vy1^2)
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook