Given,
The initial velocity of the box, u=160 m/s
The height at which the plane is flying, h=250 m
The angle at which the box is dropped, θ=0°
A)
The time of flight of a projectile motion when projected with zero launch angle is given by
[tex]t=\sqrt[]{\frac{2H}{g}}[/tex]Where g is the acceleration due to gravity and g=9.8 m/s²
On substituting the known values in the above equation,
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times250}{9.8}} \\ =7.14\text{ s} \end{gathered}[/tex]Therefore the box falls to the ground in 7.14 seconds.
B)
The range of a projectile is given by,
[tex]R=ut[/tex]On substituting the known values,
[tex]\begin{gathered} R=160\times7.14 \\ =1142.4\text{ m} \end{gathered}[/tex]Therefore the box falls to the ground at 1142.4 m away from the base camp.