# Angle to Drop a Package from Airplane

1. Dec 28, 2007

### mrlucky0

1. The problem statement, all variables and given/known data

A package is to be dropped from an airplane so that it hits the ground at a designated spot near some campers. The airplane, moving horizontally at a constant velocity of 140 km/h, approaches the spot at an altitude of .500 km above level ground. Having the designated point in sight, the pilot prepares to drop the package.

What should the angle be between the horizontal and the pilot's line of sight when the package is released?

2. Relevant equations

d = 1/2at^2
v = at

3. The attempt at a solution

I know that the velocity components of the package once it's dropped:

vx = 140 km/h
vy = -gt

The displacement components, where t is time:

dx = 140 km/h * t
dy = .50-1/2gt^2

For dy, I solved for time it takes for the package to hit the ground: .32 seconds. I also calculated that the package will travel a horizontal distance of 12.4 m in that time. Where do I go from here in determining the angle? I think I'm missing something obvious here.

Last edited: Dec 28, 2007
2. Dec 28, 2007

### Gib Z

Draw a diagram, and construct a right angled triangle such that the plane is at one of the acute angles, the campers is at the other, and the right angle is on the ground. Then on the triangle label what you know about vertical and horizontal components. Use trig!

3. Dec 28, 2007

### TVP45

Units, units, units. How high is the plane?