A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. Assume the plane is traveling horizontally with a speed of 195 km/h (54.2 m/s).
How far in advance of the recipients (horizontal distance) must the goods be dropped
time to fall
235 = .5 g t^2 = 4.9 t^2
so t = 6.93 seconds
horizontal distance = 54.2*6.93
To find the distance in advance that the goods must be dropped, we need to consider the time it takes for the supplies to reach the ground vertically, as well as the horizontal distance traveled by the plane during that time.
First, let's calculate the time it takes for the supplies to reach the ground using the vertical motion equation:
y = v0t + (1/2)at^2
Where:
y = vertical displacement (235 m)
v0 = initial vertical velocity (0 m/s, as the supplies are dropped without any initial upward or downward velocity)
t = time
a = acceleration due to gravity (-9.8 m/s^2, assuming the positive direction is upward)
Rearranging the equation, we have:
t = sqrt((2y) / a)
Plugging in the values, we get:
t = sqrt((2 * 235) / 9.8)
t = sqrt(478 / 9.8)
t ≈ 7.84 seconds
Now, since the plane is traveling horizontally, the horizontal distance traveled by the plane during this time interval will be the speed multiplied by the time:
d = v * t
Where:
d = horizontal distance
v = horizontal velocity/speed (54.2 m/s)
t = time (7.84 seconds)
Plugging in the values, we get:
d = 54.2 * 7.84
d ≈ 425.13 meters
Therefore, the goods must be dropped approximately 425.13 meters in advance of the recipients in order to reach them as they descend 235 meters vertically.