Physics
posted by Eva .
A Coast Guard airplane needs to drop a rescue package to a ship floating adrift in the ocean. The plane flies 55 m above the ocean at a speed of 215 km/h. How far away from the ship should the crew drop the package out of the plane so that the package hits the water close to the ship?

The time it will take the package to hit the ocean is T = sqrt (2H/g), where H = 55 m and g is the acceleration of gravity, 9.8 m/s^2. T = 3.35 s.
Make sure that the package is dropped a distance X ahead of the target so that
X = V T
where V is the airplane's speed in m/s. 
Alternatively:
Envision the airplane on a rescue run flying at a constant altitude "s" and with constant horizontal velocity "Vh". The objective target "O" is being observed by the pilot. When should the crew release the rescue package in order to hit the target? More specifically, at what angle to the vertical should the crew release the package?
Note that when the package is released, it retains the forward velocity of the airplane. Therefore, the package travels horizontally with a constant velocity independently of the simulaneous accelerated motion toward the earth's surface. As a result, the package will strike a point on the surface well ahead of the airplane's position at the time of release. The horizontal distance ahead of the airplane, "x", will make an angle "ø" with the vertical. Having "s", the altitude of the airplane, "x", the horizontal distance from the local vertical to the target objective "O", and "t", the time from release to the target, we have s = gt^2/2, x = Vh(t) and x = s(tanø), combining these expressions to eliminate both "x" and "t", we derive tanø = Vhsqrt(2/gs). Therefore, the crew releases the package when the pilot's instruements indicate that the angle from his local vertical position to the target is ø = arctan[Vh(sqrt(2/gs))].
Having ø, the horizontal distance from the target, at which the package should be relaeased, derives from x = stanø.