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, $g = 10\, ms^{-2}$.
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 $=t^{\prime \prime}+t+t^{\prime}=25+10+10=45 \mathrm{~s}$
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