An airplane traveling a constant 100.0 m/s is to drop supplies to flood victims isolated on a patch of land 200.0 m below. A) The supplies should be dropped how many seconds before the plane is directly overhead? B) What is the distance of the plane away from being directly overhead of the flood victims in which it should drop the package?
200 = .5gt^2
= 6.39s
To find the time at which the supplies should be dropped and the distance of the plane from being directly overhead, we can use the equations of motion.
A) To find the time at which the supplies should be dropped before the plane is directly overhead, we need to determine the time it takes for an object to fall from a height of 200.0 m. We can use the equation for the displacement of an object under free fall:
d = (1/2)gt^2
Where:
d = displacement (200.0 m)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Rearranging the equation, we get:
t^2 = (2d)/g
t = sqrt((2d)/g)
Substituting the given values, we have:
t = sqrt((2 * 200.0 m) / 9.8 m/s^2)
t ≈ sqrt(40.8 s^2)
t ≈ 6.38 s
Therefore, the supplies should be dropped approximately 6.38 seconds before the plane is directly overhead.
B) To find the distance of the plane away from being directly overhead when it should drop the package, we can determine the distance traveled by the plane during the time it takes for the supplies to fall.
The distance traveled by the plane can be calculated using the formula:
d = v * t
Where:
d = distance
v = velocity (100.0 m/s)
t = time (6.38 s)
Substituting the given values, we have:
d = 100.0 m/s * 6.38 s
d ≈ 638 m
Therefore, the plane should drop the package at a distance of approximately 638 meters away from being directly overhead of the flood victims.