Collision course of 2 objects

Question

i've tried the question in another Q&A website but nobody could really helped me there so i will try it in here. I want to calculate the point where 2 objects hit each other. More detailled: Object 1 is moving with constant velocity in a constant direction(both known). The starting point is also known. Now there is another Object 2 with constant velocity and given start position. I want to know the direction (phi) of object 2 to hit object 1. Here is a drawn example of the situation My approach to solve this i've snapshoted in the following link: Thats the equation i developed So the next step would be that i insert "t" in the equation of sin(phi), which would be very time consuming. Is that the right way? Does anybody could give me a hint for a solution?

Answer 1

Let the difference (object 2 minus object 1) of the starting positions be $\vec{\Delta}$.

Let $\hat{y}$ be a unit vector in the direction $\vec{\Delta}$, and let $\hat{x}$ be a unit vector perpendicular to $\hat{y}$ such that $\vec{v}_1$ lies in the plane spanned by $(\hat{x},\hat{y})$.

Then at the time of meeting the $x$ components are equal: $t \vec{v}_1 \cdot \hat{x} = t \vec{v}_2 \cdot \hat{x}$. This says that $$ \vec{v}_2 \cdot \hat{x} = \vec{v}_1 \cdot \hat{x} $$ And at the time of meeting the $y$ components are equal: $$ t \vec{v}_1 \cdot \hat{y} = \vec{\Delta} \cdot \hat{y} + t \vec{v}_2 \cdot \hat{y} $$ Also, the magnitude of $ \vec{v}_2 $ is known to be some value, which we can call $s_2$. So $$ \vec{v}_2 \cdot \hat{y} = \sqrt{s_2^2 - (\vec{v}_2 \cdot \hat{x} )^2} = \sqrt{s_2^2 - (\vec{v}_1 \cdot \hat{x} )^2} $$ Then, if we let $\hat{u}_2$ be the unit direction vector for object 2, so that $\vec{v}_2 =s_2\hat{u}_2$, the previous equation becomes $$ t \vec{v}_1 \cdot \hat{y} = \vec{\Delta} \cdot \hat{y} + t \sqrt{s_2^2 - (\vec{v}_1 \cdot \hat{x} )^2} $$ All the quantities in this equation are given known values, except for $t$. Solving for $t$ gives $$ t = \frac{\vec{\Delta} \cdot \hat{y}}{\vec{v}_1 \cdot \hat{y}-\sqrt{s_2^2 - (\vec{v}_1 \cdot \hat{x} )^2}} $$ Now that we know the collision time $t$, we can determine $\vec{v}_2 \cdot \hat{y}$ by $$ \vec{v}_2 \cdot \hat{y} = \frac{|\Delta|}t $$ and that determines the angle $\theta$ between $\vec{v}_2$ and the line between the starting points: $$\theta = \cos^{-1}\left( \frac{|\Delta|}{s_2t} \right) = \cos^{-1}\left( \frac{|\Delta| \left( \vec{v}_1 \cdot \hat{y}-\sqrt{s_2^2 - (\vec{v}_1 \cdot \hat{x} )^2}\right)}{s_2 (\vec{\Delta} \cdot \hat{y})} \right) = \cos^{-1}\left( \frac{ \vec{v}_1 \cdot \hat{y}-\sqrt{s_2^2 - (\vec{v}_1 \cdot \hat{x} )^2}}{s_2} \right) $$