Boats are sailing on the sea at a speed of $v$. Exactly opposite them, an airplane is flying at a height of $h$ at a speed of $V$. At what distance from the boat should the pilot drop the aid package so that it lands precisely in the boat? What would be the distance in question if the airplane is flying behind the boat; finally, if the boat is stationary on the water?
The aid package - regardless of the speed of the airplane or the boat - falls from a height of $h$ to sea level in $t=\sqrt{\frac{2 h}{g}}$ time (it is known that in free fall, the relationship between time and the distance traveled is $h=\frac{g}{2} t^{2}$).

If the airplane flies at a speed of $V$ relative to the boat, then in time $t$ the boat travels a distance of $v t$ towards the airplane, and the package, due to the inertia of the airplane's uniform motion, travels a distance of $V t$ towards the boat. The package will therefore fall into the boat if the horizontal distance between the two vehicles is

$$
s=v t+V t=(V+v) \sqrt{\frac{2 h}{g}}
$$

Using the same line of reasoning, in the case of unidirectional speed,

$$
s=(V-v) \sqrt{\frac{2 h}{g}}
$$

the package must be dropped at this distance, because at the moment of impact, the two vehicles were $(V-v) t$ apart just before the impact.

Finally, if the boat is stationary in the water, the package must be dropped at a distance of

$$
s=V \sqrt{\frac{2 h}{g}}
$$

The distance between the vehicles measured in a straight line can be calculated in all three cases using the Pythagorean theorem:

$$
d=\sqrt{s^{2}+h^{2}}
$$