I guess u could call it projectile motion... (1 Viewer)

[Sharky]

Consider a 2D-plane without the effects of gravity.

A person is standing at A(xa,ya) and another person is standing at B(xb,yb). The person at A throws a baseball with velocity Va at an angle of ATheta. The person at B fires a bullet with velocity Vb at an angle BTheta, such that the bullet collides with the baseball after t seconds at the point C(xc,yc).

Here is a summary of what we have:

KNOWNS:
=======
Positions A(xa,ya) and B(xb,yb)
Velocities Va and Vb
Angle ATheta

UNKNOWNS:
=========
Position C(xc,yc)
Time t
Angle BTheta

The question is, given that we know where both people are at and how fast their projectiles go at and the angle person A fires at, how do we find what angle person B will need to fire at and where they will subsequently collide.

There should theoretically be enough information to work it out, and it makes sense that there is only one solution.

Here's my starting contribution...

xa + Va * t * cos(ATheta) = xc
xb + Vb * t * cos(BTheta) = xc
ya + Va * t * sin(ATheta) = yc
yb + Vb * t * sin(BTheta) = yc

 

[turtle_2468]

Consider a 2D-plane without the effects of gravity.

A person is standing at A(xa,ya) and another person is standing at B(xb,yb). The person at A throws a baseball with velocity Va at an angle of ATheta. The person at B fires a bullet with velocity Vb at an angle BTheta, such that the bullet collides with the baseball after t seconds at the point C(xc,yc).

Here is a summary of what we have:

KNOWNS:
=======
Positions A(xa,ya) and B(xb,yb)
Velocities Va and Vb
Angle ATheta

UNKNOWNS:
=========
Position C(xc,yc)
Time t
Angle BTheta

The question is, given that we know where both people are at and how fast their projectiles go at and the angle person A fires at, how do we find what angle person B will need to fire at and where they will subsequently collide.

There should theoretically be enough information to work it out, and it makes sense that there is only one solution.

Here's my starting contribution...

xa + Va * t * cos(ATheta) = xc
xb + Vb * t * cos(BTheta) = xc
ya + Va * t * sin(ATheta) = yc
yb + Vb * t * sin(BTheta) = yc

The thing is, if you consider their relative motion it gets a lot easier...
ie let the first particle be defined as the origin (your euclidean plane is then a "camera" capturing what goes on about that point). Since both particles have equal acceleration, you can deem them to have zero acceleration at all.
Then consider the vector (Vb cos(BTheta)-Va cos(ATheta), Vb sin(BTheta)-Va sin(ATheta)). This vector is the velocity of B, so we just need this vector to be "aimed" at the origin from the starting point. Solve the resulting quadratic equation ((Vb sin(BTheta)-Va sin(ATheta))/(Vb cos(BTheta)-Va cos(ATheta))=yb/xb to obtain the required angle... then the distance between A and B divided by the magnitude of the vector above is the required time.

Then you can go back to the normal coord system and get position C.
In most cases, there are two solutions...

 

[Grey Council]

*cheers Turtle on*

=D

hrm, on a more serious note, we aren't supposed to know this, right? so this goes in extracurricular?

 

[grimreaper]

Im sure it could be a "harder 3u", probably q8 though