I'm asking if you know the equation to hit a moving target with a projectile.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Hard projectile physics question

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