Packages dropped by planes in the olden days didn't have onboard guidance or propulsion systems such as are found today. The path of the package was determined by the initial veloctiy (of the person that dropped that the package) and the acceleration due to gravity. If a plane were travelling at 76m/s , 4500m from the ground, how far before its target would the plan have to drop its package?

Question

Packages dropped by planes in the olden days didn't have onboard guidance or propulsion systems such as are found today. The path of the package was determined by the initial veloctiy (of the person that dropped that the package) and the acceleration due to gravity. If a plane were travelling at 76m/s , 4500m from the ground, how far before its target would the plan have to drop its package?

I don't really have an idea as where to start...I do know this though

acceleration verical=9.81 m/s^2 [down]
d vertical =4500m
veloctiy initial=76 m/s

? Help please and thanks

Answer 1

1) figure the time it takes to fall 4500 ft.

2) then, how far will something with a horizonal velocity go in that time?

Answer 2

Well I wound up figuring out my problem, before I realized you answered back...I did what you said, but why the heck would you use ft. if the question says m? I have never used ft. before! Thanks anyways!!

Answer 3

To determine how far before its target the plane would have to drop its package, we need to calculate the horizontal distance traveled by the plane during the time it takes for the package to hit the ground.

First, we need to calculate the time it takes for the package to fall from an initial height of 4500m to the ground. We can use the equation of motion:

d_vertical = (1/2) * acceleration_vertical * time^2

Rearranging the equation, we get:

time = sqrt((2 * d_vertical) / acceleration_vertical)

Substituting the given values, we have:

time = sqrt((2 * 4500m) / 9.81m/s^2)

Calculating this, we find that the time it takes for the package to fall is approximately 30.10 seconds.

Next, we need to calculate the horizontal distance traveled by the plane in 30.10 seconds. We can use the equation:

d_horizontal = velocity_horizontal * time

The velocity_horizontal of the plane is given as 76m/s, and the time is 30.10 seconds, so:

d_horizontal = 76m/s * 30.10s

Calculating this, we find that the horizontal distance traveled by the plane is approximately 2287.6 meters.

Therefore, the plane would have to drop its package approximately 2287.6 meters (or 2.29 kilometers) before its target.