A rescue pilot drops a survival kit while her plane is flying at an altitude of 1.5 km 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?

1750 meters.

First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:

d = 1/2 A T^2

where

d = distance

A = acceleration

T = time

Solving for T, gives

d = 1/2 A T^2

2d = A T^2

2d/A = T^2

sqrt (2d/A) = T

Substitute the known values and calculate.

sqrt (2d/A) = T

sqrt (2 * 1500m / 9.8 m/s^2) = T

sqrt (3000m / 9.8 m/s^2) = T

sqrt (306.122449 s^2) = T

17.49635531 s = T

Rounding to 4 significant figures gives 17.50 seconds. Since it will take

17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50

seconds before the plane goes overhead. So just simply multiply by the velocity.

17.50 s * 100 m/s = 1750 m

Given: Initial Velocity Vi = 100 m/s; Altitude y = 1.5 Km; Time t = ?

The aircraft is moving horizontally on x direction

Formula to follow: x = VicosΘt but Time t is unknown.

Solving for Time "t" formula will be y = VisineΘt + 1/2 gt²

Since sineΘt = 0 therefore y = 1/2gt² Solve for t

t = √2y/g t = √2 (1500 m) / 9.8m/s²) t = 31.24 s

Now solving for distance

x = VicosΘt x = (100 m/s) (31.24 s) x = 3,124 m or 3.124 Km