A hill is 500m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2m/s, so that its distance from the hill can be adjusted. What is the shortest time (in s) in which a packet can reach on the ground across the hill? Take g = 10m/s².

Given, height of the hill (h) = 500m u = 125 m/s
To cross the hill, the vertical component of the velocity should be sufficient to cross such height.
$\begin{array}{l}\therefore {\mathrm{u}}_{\mathrm{y}}\ge \sqrt{2\mathrm{gh}}\\ \ge \sqrt{2\times 10\times 500}\\ \ge 100\mathrm{m}/\mathrm{s}\\ \text{ But }{\mathrm{u}}^2={\mathrm{u}}_{\mathrm{x}}^2+{\mathrm{u}}_{\mathrm{y}}^2\end{array}$

Horizontal component of initial velocity
$\begin{array}{l}{\mathrm{u}}_{\mathrm{x}}=\sqrt{{\mathrm{u}}^2-{\mathrm{u}}_{\mathrm{y}}^2}\\ =\sqrt{(125{)}^2-(100{)}^2}=75\mathrm{m}/\mathrm{s}\end{array}$
Time taken to reach the top of the hill
$\mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}=\sqrt{\frac{2\times 500}{10}}=10\mathrm{s}$
Time taken to reach the ground from the top if the hill t' = t = 10s
Horizontal distance travelled in 10s.
x = u_x x t
= 75 x 10 = 750m
Distance through which canon has to be moved = 800 = 750 = 50m
Speed with which canon can move = 2m/s
Time taken by canon = 50/2
t=25s
Total time taken by a packet to reach on the ground
= t'' + t + t'
= 25 + 10 + 10 = 45s