Hard projectile physics question

In summary, the conversation discusses the equation needed to hit a moving target with a projectile. The canon is located at (0,0) and the target is at (d,0), moving away with a constant x speed of vx2. The canon's angle of elevation is represented by a, and the question is to solve for the power of the cannon, v. Using basic projectile physics, the required y velocity of the projectile is found to be g*d/(2*(vx1 - vx2)), where vx1 and vx2 are the initial and target x velocities respectively. To find the launch speed and angle, the equations vx1=v*cos(a) and vy1=v*sin(a) are used, and the final equation for v
  • #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?
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  • #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

1. What is a hard projectile?

A hard projectile is any object that is propelled through the air or space by a force, such as a bullet, cannonball, or spacecraft.

2. What is projectile motion?

Projectile motion is the curved path that a hard projectile follows as it is propelled through the air by a force. This is influenced by the initial velocity, angle of launch, and the effects of gravity.

3. How is the trajectory of a projectile calculated?

The trajectory of a projectile can be calculated using the equations of motion, which take into account the initial velocity, angle of launch, and acceleration due to gravity. These calculations can also be aided by computer simulations.

4. What is the difference between a hard projectile and a soft projectile?

A hard projectile is typically solid and maintains its shape during flight, while a soft projectile is deformable and may change shape during flight. Hard projectiles are often used for long-range and precise targeting, while soft projectiles are used for shorter distances and impact damage.

5. How does air resistance affect a projectile?

Air resistance, also known as drag, slows down the velocity of a projectile and can cause it to deviate from its intended trajectory. This effect is more significant for larger and slower projectiles, and it can be reduced by using streamlined shapes or increasing the initial velocity.

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