$A$ hill is $500 \, m$ high. Supplies are to be sent across the hill,using a cannon that can hurl packets at a speed of $125 \, m/s$ over the hill. The cannon is located at a distance of $800 \, m$ from the foot of the hill and can be moved on the ground at a speed of $2 \, m/s$; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach the ground on the other side of the hill? Take $g = 10 \, m/s^2$.

(D) Given: Speed of packets $u = 125 \, m/s$,Height of hill $h = 500 \, m$,Acceleration due to gravity $g = 10 \, m/s^2$.
To just clear the hill,the vertical component of velocity $u_y$ must satisfy $u_y^2 = 2gh$.
$u_y = \sqrt{2 \times 10 \times 500} = 100 \, m/s$.
The horizontal component of velocity is $u_x = \sqrt{u^2 - u_y^2} = \sqrt{125^2 - 100^2} = \sqrt{15625 - 10000} = \sqrt{5625} = 75 \, m/s$.
Time taken to reach the top of the hill is $t_1 = u_y / g = 100 / 10 = 10 \, s$.
Time taken to fall from the top to the ground is $t_2 = \sqrt{2h/g} = \sqrt{2 \times 500 / 10} = 10 \, s$.
Total time of flight $T = t_1 + t_2 = 10 + 10 = 20 \, s$.
Horizontal distance covered by the packet during flight is $x = u_x \times T = 75 \times 20 = 1500 \, m$.
Since the cannon is initially $800 \, m$ from the hill,and the packet travels $1500 \, m$ horizontally,the cannon must be moved to a position such that the packet lands just on the other side of the hill.
The required horizontal distance from the hill is $x_{req} = 1500 \, m$.
The cannon is currently $800 \, m$ away. To minimize time,we move the cannon towards the hill. The distance to move is $d = 800 - (1500 - 800) = 100 \, m$ (if we consider the landing point relative to the hill).
Actually,the cannon must be at a distance $x_{req} = 1500 \, m$ from the landing point. Since it is $800 \, m$ from the hill,it needs to be moved $700 \, m$ closer to the hill.
Time to move $t_{move} = 700 / 2 = 350 \, s$.
Total time $= t_{move} + T = 350 + 20 = 370 \, s$.