Kinematics Horizontal Motion Problem

Question

A rescue pilot drops a survival kit while her plane is flying horizontally at 117 m/s from an altitude of 2.35 x 10(cubed) m above the ground. If air friction is ignored, how far horizontally in advance of the starving explorers drop zone should she release the package?

Answer 1

First, consider only the vertical drop. The fact that the kit is moving horizontally does not affect its "vertical drop" profile. It drops the same vertically as if it were dropped from a stationary position.

Vertically, we can say that y(t) = 1/2at² where "y" is the height, "a" is the acceleration due to gravity (9.81 m/s²) and t is the time. So:

2.35 x 10³ =(1/2)(9.81)(t²)

t² = (2.35 x 10³)(2)/9.81

t² = 479.103

t = √479.103 = 21.888 seconds.

Now, consider the horizontal travel. It is travelling at 117 m/s and since we are ignoring air friction, it will maintain that speed until it hits the ground (horizontal direction only - what's happening vertically does not change what is happening horizontally). velocity = distance/time so distance (horizontal distance) = velocity • time

d = 117•21.888 = 2560.945 m

The "givens" have 3 significant figures so we round our answer to 3 sig figs. She must release the kit 2560 meters prior to drop zone.