(a) 278.4 m
First of all, we need to find the time it takes for the package to reach the ground. We can use the equation for the vertical motion:
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
where
h = -165 m is the vertical displacement
[tex]u_y = 0[/tex] is the initial vertical velocity
[tex]g=-9.8 m/s^2[/tex] is the acceleration of gravity
t is the time
Substituting and solving for t,
[tex]t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(-165)}{9.8}}=5.80 s[/tex]
During this time interval, the package travels with a horizontal speed of
[tex]v_x = 48 m/s[/tex] (the initial velocity of the plane)
So, the horizontal distance covered is
[tex]d=v_x t = (48)(5.80)=278.4 m[/tex]
So, the package landed 278.4 m far from the point directly below the releasing point.
(b) 48 m/s
Along the horizontal direction, there is no force (if we neglect air resistance), therefore no acceleration, according to Newton's second law:
[tex]\sum F = ma[/tex]
so a = 0 since [tex]\sum F = 0[/tex].
This means that the horizontal component of the velocity remains constant; since its initial value is
[tex]v_x = 48 m/s[/tex]
Then it will remain constant for the whole motion, and therefore it will still be 48 m/s when the package reaches the ground.
