A rescue pilot drops a survival kit while her plane is flying at an altitude of 2000.0 m
with a forward velocity of 100.0 m/s. If air friction is disregarded, how far in advance
of the starving explorer’s drop zone should she release the package?
6 Answers

A rescue pilot drops a survival kit while her plane is flying at an altitude of 2000.0 m with a forward velocity of 100.0 m/s. If air friction is disregarded, how far in advance of the starving explorer’s drop zone should she release the package?
The equation for a falling object
Distance = (Initial velocity * time) + (½ * 9.8 * time^2)
Distance = 2000.0 m
Initial velocity = 0 m/s
2000.0 = (0 * time) + (4.9 * time^2)
2000.0 = 4.9 * t^2
t = 20.2 seconds
As the survival kit drops vertically for 20.2 seconds, it moves horizontally for 20.2 seconds at a velocity of 100.0 m/s.
Horizontal distance = 100.0 m/s * 20.2 seconds = 2020 meters
As the survival kit drops for 20.2 seconds, it moves horizontally 2020 meters.
How far in advance of the starving explorer’s drop zone should she release the package? 2020 meters

Need to figure out how long it would take for the object to fall straight down from 2000 m with 0 initial velocity and only gravity affecting it.
Then, use that time and the 100 m/s to figure out the horizontal distance it will travel during the fall.

Velocity Kit Plane

Get Survival Things You Lost!

When the package is released it will continue for 2,019 meters. So drop it that far from the target.
Easy equation. Given
H = 2,000 m
Vo = 100 m/s
g = 9.81 m/s^2
R = Vo * SQRT { [2H] / g }
R = (100 m/s) * SQRT { [ 2 * (2,000 m) ] / (9.81 m/s^2) }
R = (100 m/s) * SQRT { [ 4,000 m ] / (9.81 m/s^2) }
R = (100 m/s) * SQRT { 407.7 m^2/s^2 }
R = (100 m/s) * (20.2 m/s)
R = 2,019 m

I can't really remember how to do this, but you have to use this knematic equation: vf^2 = vi^2 + 2ad
vf (final velocity) = 0
vi (initial velocity) = 100
a = 9.8 (acceleration due to gravity)
d = ?