The pilot must drop the pennant on the deck of a small vessel ahead of the aircraft.

The pilot must drop the pennant on the deck of a small vessel ahead of the aircraft. The airplane flies horizontally at a height h with a constant speed v, the ship moves towards the airplane with a speed u. Find the distance (in a straight line) between the ship and the aircraft at the moment the pennant is dropped, if the pennant is known to hit the target. Determine the fall time of the pennant.

Given:

h is the altitude at which the plane is flying;

v is the aircraft speed;

u is the speed of the vessel moving towards the aircraft;

g – acceleration of gravity (9.8 m / s ^ 2).

It is required to determine the distance from the aircraft to the ship S, necessary for the pennant dropped from the aircraft to hit the ship, as well as the time t of the pennant falling.

Since the plane flies horizontally, the pennant dropped from it will reach the ground in the time:

t = (2 * h / g) ^ 0.5.

Since the aircraft and the ship are moving towards each other, the speed of the aircraft relative to the ship will be equal to:

v1 = v + u.

It turns out that in order for the pennant to hit the deck of the ship, the plane must drop it from a distance to the ship:

S = v1 * t = (v + u) * (2 * h / g) ^ 0.5.

Answer: the distance is (v + u) * (2 * h / g) ^ 0.5, the fall time of the pennant is (2 * h / g) ^ 0.5.