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:
[tex]s=ut+\frac{1}{2}at^2[/tex]
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
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find the time at which the package reaches the ground (time of flight):
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(150)}{9.8}}=5.53 s[/tex]
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
[tex]d=v_x t[/tex]
where
[tex]v_x=43 m/s[/tex] is the horizontal velocity of the package
t = 5.53 s is the time of flight
Solving for d,
[tex]d=(43)(5.53)=237.8 m[/tex]
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
[tex]v=u+at[/tex]
where
u = 0
[tex]a=g=-9.8 m/s^2[/tex] 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,
[tex]v=0+(-9.8)(5.53)=-54.2 m/s[/tex]
