We're given two particles $p_1$ and $p_2$ with time-dependent position vectors, $\mathbf{r}_{1}$ & $\mathbf{r}_{2}$ ($u_0, v_0$ and $g$ constant):
$$\mathbf{r}_{1}\left(t\right) = \mathbf{j} u_{0}t – \mathbf{k} \frac{gt^2}{2}$$
$$\mathbf{r}_{2}\left(t\right)=\mathbf{i}v_{0}t – \mathbf{k}\frac{gt^2}{2}$$
We're asked to find the angle $\theta$ between the vectors at times $t=0$ and $t \to\infty$.
Hint given: figure out which vector points along the path.
The velocity vectors point along the path of motion of each particle, so I first found them by differentiation:
$$\mathbf{V}_1\left(t\right) = \mathbf{j}u_0 – \mathbf{k}gt$$
$$\mathbf{V}_2\left(t\right) = \mathbf{i}v_0 – \mathbf{k}gt$$
Now it makes sense to get the angle between them using a rearrangement of the dot product for vectors:
$$\cos{\theta}=\frac{\mathbf{\mathbf{V_1}}\cdot\mathbf{V_2}}{\left|\mathbf{V}_1\right|\cdot\left|\mathbf{V}_2\right|}$$
This works fine for $t=0$ , and $\theta$ = $\frac{\pi}{2}$.
I attempted a solution for $t \to \infty$ by getting limits for each component of each vector, which is fine for the first component in both cases as they are constant. The $\mathbf{k}$ component in both cases, however, is giving me trouble, as I see no other result than $\infty$ which appears to be unworkable.
I do happen to know that for $t \to \infty$ , $\cos{\theta}$ should equal 1. Some help to get to this fact would be appreciated.
Aside: Please let me know if my vector formatting in LaTeX could be improved, I'm kinda new to it.
Best Answer
We can see that $$\mathbf V_1\cdot\mathbf V_2 = g^2t^2$$ Also $$\vert \mathbf V_1\vert=\sqrt{u_0^2+g^2t^2}, \ \vert\mathbf V_2\vert=\sqrt{v_0^2+g^2t^2}$$
So we have $$\begin{align}\cos \theta&=\frac{g^2t^2}{\sqrt{u_0^2+g^2t^2}\sqrt{v_0^2+g^2t^2}}\\&=\frac{g^2}{\sqrt{\frac{u_0^2}{t^2}+g^2}\sqrt{\frac{v_0^2}{t^2}+g^2}}\end{align}$$ So at $t\rightarrow\infty$, we have $$\cos \theta=\frac{g^2}{\sqrt{g^2}\sqrt{g^2}}=1$$ OR $$\theta = 0$$