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
[tex]y(t) = h + u_y t + \frac{1}{2}gt^2[/tex]
where
h = 117 m is the initial height
[tex]u_y = 0[/tex] 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:
[tex]0=h+\frac{1}{2}gt^2\\t=\sqrt{-\frac{2h}{g}}=\sqrt{-\frac{2(117)}{-9.8}}=4.89 s[/tex]
Now we can find the horizontal distance travelled by the package by considering the horizontal motion only; so it is given by
[tex]d=v_x t[/tex]
where
[tex]v_x = 42 m/s[/tex] is the horizontal velocity of the package (which is constant)
t = 4.89 s
Substituting,
[tex]d=(42)(4.89)=205 m[/tex]
So, the package lands 205 m ahead of the point directly below the plane where the package was released.
