Question

An Alaskan rescue plane traveling 42 m/s

Answer 1

Answer:

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

$y(t) = h + u_y t + \frac{1}{2}gt^2$

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:

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

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.