Question

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 ms^-1over 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, 9 = 10 ms^-2) [NCERT Exemplar]

Difficult

Solution

Given, speed of packets = 125 ms^-1
Height of the hill = 500 m
To cross the hill, the vertical component of the velocity should be sufficient to cross such height.
$u_{y} \geq \sqrt{2 g h}$
$\geq \sqrt{2 \times 10 \times 500} \geq 100 {ms}^{- 1}$
But $u^{2} = u_{x}^{2} + u_{y}^{2}$
$∴$ Horizontal component of initial velocity
$u_{x} = \sqrt{u^{2} - u_{y}^{2}}$
$= \sqrt{(125)^{2} - (100)^{2}}$
$= 75 {ms}^{- 1}$
Time taken to reach the top of the hill.
$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 500}{10}} = 10 s$
Time taken to reach the ground from the top of the hill.
$t^{'} = t = 10 s$
Horizontal distance travelled in 10 s,
$x = u_{x} \times t = 75 \times 10$
=750m
$∴$ Distance through which canon has to be moved
= 800 - 750 = 50 m
Speed with which canon can move = 2 ms^-1
$∴$ Time taken by canon. $ℓ^{'} = \frac{50}{2}$
$\Rightarrow$ = 25s
$∴$ Total time taken by a packet to reach on the ground
$= t^{''} + t + t^{'}$
= 25 + 10 + 10 = 45 s

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