You have a position $u(t_0)$ of your moving object at current time $t_0$ and part of the change of the position (which is the velocity $\dot{u}(t_0)$) at current time. The part you have is the direction of the velocity vector, you lack it‘s magnitude. E.g.
$$
\dot{u}(t_0) = \lVert \dot{u}(t_0) \rVert \, (\sin \alpha(t_0), \cos \alpha(t_0))^\top
$$
where $\alpha$ is the angle to the $y$-axis.
To calculate the path of the moving object from an initial position and velocities over time
$$
u(t) = \int\limits_{t_0}^t \dot{u}(\tau) \, d\tau
$$
this is not enough information.
So you must add assumptions.
E.g. the direction will not change over time $\alpha(t) = \alpha$ and there will be non-zero magnitudes (it moves forward somehow).
This would imply that the path will be a straight line (neglecting Earth‘s surface curvature, otherwise one would need to deal with parts of circle arcs, neglecting the deviations of Earth‘s surface from an ideal sphere, yadda, yadda).
In this simplified scenario calculating the distances of the target points to the line and picking the one with smaller distance would be the solution.
This would also need a check that points lie in forward direction relative to initial position and velocity direction.
The line can be modeled as
\begin{align}
n \cdot u &= d \iff \\
(n_x, n_y) \ (x, y)^\top &= n_x x + n_y y = d
\end{align}
with a unit normal vector $n$ of the line, a position vector $u$ and the distance $d$ of the line to the origin. $\top$ is short for transposition, it turns a column vector into a row vector and vice versa (ignore it, if you do not know it).
A target $u_i = (x_i, y_i)^\top$ will lie on a parallel line
$$
n \cdot u_i = d_i
$$
where the resulting number $d_i$ is the distance of the parallel line to the origin.
The distance of $u_i$ to the path line then is the absolute difference
$$
\lvert d-d_i\rvert
$$.
We start with the path line represented as
\begin{align}
u(t) &= u_0 + v t \\
u(t) &= (x(t), y(t))^\top \\
u_0 &= u(t_0) \\
v &= (\sin \alpha, \cos \alpha)^\top
\end{align}
then a unit normal vector to that line is
$$
n = (–\cos \alpha, \sin \alpha)^\top
$$
A target $u_i$ lies in forward direction if
\begin{align}
a_i = v \cdot (u_i - u_0)
&= \lVert v \rVert \, \lVert u_i - u_0 \rVert \cos \angle(v, u_i-u_0) \\
&= (\sin \alpha, \cos \alpha) \, (x_i-x_0, y_i-y_0)^\top \\
&= (x_i-x_0) \sin \alpha + (y_i-y_0) \cos \alpha > 0
\end{align}
Example:
(Large version)
You can fiddle with this example here: link
Play with the sliders to vary the various parameters of the scene.
