Question

An Alaskan rescue plane traveling 48 m/s drops a package of emergency rations from a height of 165 m to a stranded party of explorers. The acceleration of gravity is 9.8 m/s 2 . Where does the package strike the ground relative to the point directly below where it was released? Answer in units of m.

What is the horizontal component of the velocity just before it hits?
Answer in units of m/s.​

Answer 1

(a) 278.4 m

First of all, we need to find the time it takes for the package to reach the ground. We can use the equation for the vertical motion:

`h=u_yt+(1)/(2)gt^2`

where

h = -165 m is the vertical displacement

`u_y = 0` is the initial vertical velocity

`g=-9.8 m/s^2` is the acceleration of gravity

t is the time

Substituting and solving for t,

`t=\sqrt{(2h)/(g)}=\sqrt{(2(-165))/(9.8)}=5.80 s`

During this time interval, the package travels with a horizontal speed of

`v_x = 48 m/s` (the initial velocity of the plane)

So, the horizontal distance covered is

`d=v_x t = (48)(5.80)=278.4 m`

So, the package landed 278.4 m far from the point directly below the releasing point.

(b) 48 m/s

Along the horizontal direction, there is no force (if we neglect air resistance), therefore no acceleration, according to Newton's second law:

`\sum F = ma`

so a = 0 since
`\sum F = 0`.

This means that the horizontal component of the velocity remains constant; since its initial value is

`v_x = 48 m/s`

Then it will remain constant for the whole motion, and therefore it will still be 48 m/s when the package reaches the ground.