Question

A plane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 meters must drop an emergency packet on a target on the ground. the trajectory of the packet is given by x=100t, y=-4.9t^2+4000, for t≥0, where the origin is the point on the ground beneath the plane at the moment of release.

A: how many horizontal meters(to the nearest tenth of a meter) from the target should the packet be released in order to hit the target?
B: how far (to the nearest tenth of a meter) does the package travel from the moment of release until the time it hits the target?

Answer 1

A. 2857.1 meters

B. 4915.6 meters

Explanation:

A.

First we need to find when the packet will hit the surface level, and we do this by finding the value of t when y = 0:

0 = -4.9t^2 + 4000

4.9t^2 = 4000

t^2 = 816.33

t = 28.571 s

Then, we use this time in the horizontal distance equation to find the value of x:

x = 100 * 28.571 = 2857.1 meters

B.

To find the distance between the inicial and final location of the package, we can use the Pythagoras' theorem with the horizontal and vertical distance traveled:

distance^2 = x^2 + y^2

Using x = 2857.1 and y = 4000, we have:

distance^2 = 2857.1^2 + 4000^2

distance^2 = 24163020.41

distance = 4915.6 meters