Because the coconut and the projectile both fall at the same rate,

the position of the coconut relative to the projectile is independent of the the acceleration due to gravity, \(\vec{a}=-g\hat{j}\).

Let \(\vec{r}_c(t)\) and \(\vec{r}_p(t)\) be the position vectors of the coconut and projectile, respectively,

and take the origin to be the position of the projectile at time \(t=0\).

The projectile is fired with a velocity \(\vec{v}_0\) in the direction of the coconut at time \(t=0\), namely in the same direction as the vector \(\vec{r}_{c0}\).

In the absence of gravity (\(\vec{a}=\vec{0}\)), \(\vec{r}_c(t)=\vec{r}_{c0}\) and \(\vec{r}_p(t)=\vec{v}_0 t\), so the projectile follows a straight line path in the direction of \(\vec{v}_0\) towards \(\vec{r}_{c0}\).

In that case the position vector of the coconut relative to the projectile at time \(t\) is \(\Delta \vec{r}(t) = \vec{r}_c(t)-\vec{r}_p(t)=\vec{r}_{c0}-\vec{v}_0 t\).

Including gravity, the position vector of the coconut at time \(t\) is \(\vec{r}_c(t) = \vec{r}_{c0}+\frac{1}{2} \vec{a} t^2\),

and the position vector of the projectile at time \(t\) is \(\vec{r}_p(t) = \vec{v}_0 t+\frac{1}{2} \vec{a} t^2\).

Therefore the position vector of the coconut relative to the projectile at time \(t\) is \(\Delta \vec{r}(t) = \vec{r}_c(t)-\vec{r}_p(t)=\vec{r}_{c0}-\vec{v}_0 t\),

which is identical to the relative position in the absence of \(\vec{a}\)!

When you run the animation, notice that the vector \(\Delta \vec{r}(t)\) connecting the projectile to the coconut

remains parallel to the original straight line path, and continues to shrink with time until it vanishes.