Question

An Alaskan rescue plane traveling 42 m/s

drops a package of emergency rations from
a height of 117 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

Answer 1

205 m

Step-by-step explanation:

We can find the time of flight of the package by considering its vertical motion first. In fact, the vertical position at time t is given by

`y(t) = h + u_y t + (1)/(2)gt^2`

where

h = 117 m is the initial height

`u_y = 0` is the initial vertical velocity

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

The package reaches the ground when y=0, so substituting this, we find the corresponding time:

`0=h+(1)/(2)gt^2\\t=\sqrt{-(2h)/(g)}=\sqrt{-(2(117))/(-9.8)}=4.89 s`

Now we can find the horizontal distance travelled by the package by considering the horizontal motion only; so it is given by

`d=v_x t`

where

`v_x = 42 m/s` is the horizontal velocity of the package (which is constant)

t = 4.89 s

Substituting,

`d=(42)(4.89)=205 m`

So, the package lands 205 m ahead of the point directly below the plane where the package was released.