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
![t=\sqrt[]{(2H)/(g)}](https://img.qammunity.org/2023/formulas/physics/college/ae4qm1zs6b2mg5rxdyawp2o6w4ov9r6jn5.png)
Where g is the acceleration due to gravity and g=9.8 m/s²
On substituting the known values in the above equation,
![\begin{gathered} t=\sqrt[]{(2*250)/(9.8)} \\ =7.14\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/2cqwtxogzf5885l42tualen1cr9rgvylu6.png)
Therefore the box falls to the ground in 7.14 seconds.
B)
The range of a projectile is given by,

On substituting the known values,

Therefore the box falls to the ground at 1142.4 m away from the base camp.