A hill is $500\, m$ 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 $800 \,m$ from the foot of hill and can be moved on the ground at a speed of $2\, ms^{-1}$; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take, $g = 10\, ms^{-2}$.

Given, speed of packets $=125 \mathrm{~m} / \mathrm{s}$

Height of the hill $=500 \mathrm{~m}$

To cross the hill, the vertical component of the velocity should be sufficient to cross such height.

$\mathrm{U}_{y} \geq \sqrt{2 g h}$

$\geq \sqrt{2 \times 10 \times 500}$

$\geq 100 \mathrm{~m} / \mathrm{s}$

But $\mathrm{U}^{2}=\mathrm{U}_{x}^{2}+\mathrm{U}_{y}^{2}$

$\therefore$ Horizontal component of initial velocity,

$\mathrm{U}_{x}=\sqrt{\mathrm{U}^{2}-\mathrm{U}_{y}^{2}}=\sqrt{(125)^{2}-(100)^{2}}=75 \mathrm{~m} /\mathrm{s}$

Time taken to reach the top of the hill,

$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=10 \mathrm{~s}$

Time taken to reach the ground from the top of the hill $t^{\prime}=t 10 \mathrm{~s}$. Horizontal distance travelled in $10 \mathrm{~s}$.

$x=\mathrm{U}_{x} \times t=75 \times 10=750 \mathrm{~m}$

$\therefore$ Distance through which canon has to be moved $=800-750=50 \mathrm{~m}$

Speed with which canon can move $=2 \mathrm{~m} / \mathrm{s}$

$\therefore$ Time taken by canon $=\frac{50}{2} \Rightarrow t^{\prime \prime}=25 \mathrm{~s}$

$\therefore$ Total time taken by a packet to reach on the ground $=t^{\prime \prime}+t+t^{\prime}=25+10+10=45 \mathrm{~s}$

A gun can fire shells with maximum speed $v_{0}$ and the maximum horizontal range that can be achieved is $\mathrm{R}=\frac{v_{0}^{2}}{g}$. If a target farther away by distance $\Delta x$ (beyond $\mathrm{R}$ ) has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least