A rescue pilot drops a survival kit while her plane is flying at an altitude of 1200.0 meters with a forward velocity of 97.4 m/s. How far in advance of the drop zone should she release the package?

Answer 1

When the pilot drops the kit, the kit continues at the speed of the plane.

So the answer to the question is the horizontal distance the package flies before hitting the ground.

I am assuming that we are to ignore air drag.

The kit follows a parabola resulting from a combination of two movements:

- A horizontal trajectory due to inertia, which is what the kit would

follow in the absence of gravity.

- A vertical fall due to gravity, which is what the kit would follow

in the absence of any initial horizontal velocity.

You can view the trajectory in an X-Y coordinate space, where

the X coordinate is along the ground, and

the Y coordinate is vertical right where the kit gets released.

Then [x(t), y(t)] is the position of the kit after time t.

(We start measuring time t from the instant the kit leaves the plane.)

x(t) = vt

y(t) = h - gt²/2,

where

v is the initial horizontal velocity

h is the height of the plane

g is the gravitational acceleration

The fall continues until the kit hits the ground at some time t₁,

which we can calculate, because at time t₁, y(t₁) = 0. That is,

h - gt₁²/2 = 0

t₁ = √(2h/g).

Having obtained t₁, we can calculate how far the kit got horizontally by:

x(t₁) = vt₁ = v√(2h/g).

Substituting

h = 1200m,

v = 97.4m/s

g = 9.8m/s²

x(t₁) = 97.4m/s √(2 x 1200m / 9.8 m/s²) = 1524.2 m