In space you don't just "go somewhere".
You have to match orbits, while not wasting too much fuel.
If you're in a low circular orbit, and you want to get to a high circular orbit, it takes two tangential burns, one to elongate your orbit into an ellipse, and another at the high point of the ellipse to make it circular again.
This is called a Hohman transfer.
You may have to do this multiple times, depending on how much thrust you have.
If your orbit is in a different plane from the orbit of the space station, you have to wait until you reach the plane of the other orbit, then do a lateral burn. You may have to do this several times to change your orbit's angle sufficiently, each time having to wait another half-orbit.
EDIT: to give some perspective on this, if your orbit crosses the plane of the other orbit at an angle of 10 degrees, that means you are crossing that plane at about one mile per second.
(Orbit velocity times sin(10 degrees).)
If your rocket motor generates 1G of thrust, you need to run it around 2.5 minutes to get aligned with that plane. (5280/32/60)
REVISED:
If you're in the same orbit as your destination, but some distance behind it (say), the way you catch up is by getting into a lower orbit by a Hohman transfer, with greater angular velocity, and then another such transfer to get back to the original orbit.
This is called orbit phasing.
If you just accelerate toward the object, that would put you in an orbit that rises above the target, and then eventually falls further behind because it is a higher orbit.
If the stick is a bar, having plain contact with the ground along its length, the friction force opposing the rotation suggests to model it as 2 cantilever beams with uniformly distributed load, fixed in the COM. The friction load is distributed along its length, resulting in max. torque close to COM and zero at the ends.
So for a small area close to the ends, the total torque results only from the load on this area:
$$\delta \tau = \delta I\frac{d\omega}{dt}$$
$\delta \tau = \delta Fr$ and the friction force in the element is $\delta F = \mu \delta N = \mu \delta m g$
The moment of inertia $\delta I = \delta m r^2$ and $$\omega = \frac{v}{r}$$
So,
$$\mu \delta m g r = \delta m r^2 \frac{1}{r} \frac{dv}{dt} \implies \frac{dv}{dt} = \mu g $$
If we elaluate the force to decrease the average translation velocity in the same region: $$\delta F = \delta m\frac{dv_t}{dt} = \mu \delta N = \mu \delta mg \implies \frac{dv_t}{dt} = \mu g$$
Under the same acceleration, they must decrease together. If it happens for the ends of the bar, all the body will stop spinning and moving linearly at the same time for this model.
But if for example, the central portion have contact but not the ends, it is perfectly possible for the bar keeps rotating, after stopping its translational movement.
Best Answer
I would say that it is a result of time reversal symmetry. If you consider the projectile at the apex of its trajectory then all that changes under time reversal is the direction of the horizontal component of motion. This means that the trajectory of the particle to get to that point and its trajectory after that point should be identical apart from a mirror inversion.