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

205 m

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

where

h = 117 m is the initial height

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:

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

where

is the horizontal velocity of the package (which is constant)

t = 4.89 s

Substituting,

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

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be