Question

PART ONE: An Alaskan rescue plane traveling 43 m/s

drops a package of emergency rations from
a height of 150 m to a stranded party of explorers.
The acceleration of gravity is 9.8 m/s
Where does the package strike the ground
relative to the point directly below where it
was released?
Answer in units of m.
PART TWO: What is the horizontal component of the velocity just before it hits?
Answer in units of m/s
PART THREE: What is the vertical component of the velocity
just before it hits? (Choose upward as the
positive vertical direction)
Answer in units of m/s.

Answer 1

1) 237.8 m

We start by considering the vertical motion of the package. This is a free fall motion (uniform accelerated motion). For this, we can use the suvat equation:

`s=ut+(1)/(2)at^2`

where

, chosing downward as positive direction,

s = 150 m is the vertical displacement

u = 0 is the initial vertical velocity of the package

t is the time

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

Solving for t, we find the time at which the package reaches the ground (time of flight):

`t=\sqrt{(2s)/(g)}=\sqrt{(2(150))/(9.8)}=5.53 s`

Let's now consider the horizontal motion of the package: this is a uniform motion, since the package mantains its initial horizontal velocity (the same as the plane). The horizontal distance travelled is given by the package is

`d=v_x t`

where

`v_x=43 m/s` is the horizontal velocity of the package

t = 5.53 s is the time of flight

Solving for d,

`d=(43)(5.53)=237.8 m`

Therefore, the package lands 237.8 m away from the point directly below where it was released.

2) 43 m/s

The motion of the package along the horizontal direction is a uniform motion: it means that the horizontal component of the velocity is constant. Since its initial horizontal velocity was equal to the velocity of the plane, 43 m/s, it means that the final component of the horizontal velocity is exactly the same, 43 m/s.

3) -54.2 m/s

The vertical component of the velocity of the package just before hitting the ground can be calculated using the suvat equation

`v=u+at`

where

u = 0

`a=g=-9.8 m/s^2` is the acceleration of gravity (here we have chosen upward as positive direction, so the acceleration is negative, since it is downward)

t = 5.53 s is the time of flight

Substituting,

`v=0+(-9.8)(5.53)=-54.2 m/s`