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?