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.
To determine the distance at which the crew should drop the rescue package, we need to consider the time it takes for the package to reach the ship and the horizontal distance covered by the plane in that time.
Since the plane is flying at a constant speed of 215 km/h, we can say that the distance covered by the plane is equal to its speed multiplied by the time it takes to reach the ship.
To find the time it takes for the package to reach the ship, we need to use the concept of time.
The formula for time is distance divided by speed:
time = distance / speed
Given that the plane flies 55 meters above the ocean, we need to find the horizontal distance that the plane covers. To do this, we can use the Pythagorean theorem.
The Pythagorean theorem states that the square of the hypotenuse (distance covered by the plane) is equal to the sum of the squares of the two other sides (distance above the ocean and distance from the plane to the ship).
Using this, we have:
(distance covered by the plane)^2 = (distance above the ocean)^2 + (distance from the plane to the ship)^2
Let's solve it step by step:
(distance covered by the plane)^2 = (55)^2 + (distance from the plane to the ship)^2
(distance covered by the plane)^2 = 3025 + (distance from the plane to the ship)^2
Since the plane flies at a constant speed and the time to reach the ship is the same for both the vertical distance and horizontal distance, we can now equate the time equation and the distance equation.
(distance covered by the plane)^2 = 3025 + (distance from the plane to the ship)^2
(distance covered by the plane)^2 = (215 km/h * time)^2
Since we have the speed of the plane in km/h, we need to convert it to m/s to ensure consistent units. 1 km/h = 0.2778 m/s.
distance covered by the plane = speed * time
(distance covered by the plane)^2 = (215 km/h * 0.2778 m/s * time)^2
(distance covered by the plane)^2 = (59.5 m/s * time)^2
Equating the two equations, we can solve for time:
(59.5 m/s * time)^2 = 3025 + (distance from the plane to the ship)^2
Now let's isolate the distance from the plane to the ship:
(distance from the plane to the ship)^2 = (59.5 m/s * time)^2 - 3025
(distance from the plane to the ship)^2 = (59.5 * time)^2 - 3025
We do not have the time yet, so let's find it using the formula:
time = distance / speed
The distance that the package needs to cover vertically is from the plane to the ship, which is 55 meters.
time = 55 m / 0.2778 m/s (converting km/h to m/s)
time = 198.1 seconds (approx.)
Now we can plug this time value into the equation:
(distance from the plane to the ship)^2 = (59.5 * 198.1)^2 - 3025
(distance from the plane to the ship)^2 = 235157.81
Taking the square root of both sides, we find:
distance from the plane to the ship ≈ √235157.81
distance from the plane to the ship ≈ 485.0 meters (rounded to the nearest meter)
Therefore, the crew should drop the rescue package approximately 485 meters away from the ship, horizontally, to ensure it lands close to the ship in the ocean.