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 $125 \, m \, s^{- 1}$ 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 $2 \, m \, s^{- 1}$ so that its distance from the hill can be adjusted. What is the shortest time (in seconds) in which a packet can reach on the ground across the hill? (Take $g=10 \, m \, s^{- 2}$ )

Answer 1

Given, height of the hill $(h)=500\, m$
$u=125\, m / s$
To cross the hill, the vertical component of the velocity should be sufficient to cross
such height.
$\therefore u _{ y } \geq \sqrt{2 gh }$
$\geq \sqrt{2 \times 10 \times 500}$
$\geq 100 \,m / s$
But $u ^{2}= u _{ x }^{2}+ u _{ y }^{2}$
$\therefore $ Horizontal component of initial velocity
$u _{ x }=\sqrt{ u ^{2}- u _{ y }^{2}}$
$=\sqrt{(125)^{2}-(100)^{2}}$
$=75 \,m / s$
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 $t ^{\prime}= t =10\, s$
Horizontal distance travelled in $10\, s$
$ x = u _{ x } \times t $
$=75 \times 10 $
$=750\, m $
$\therefore $ Distance through which canon has to be moved
$=800-750 $
$=50\, m$
Speed with which canon can move $=2\, m / s$
$\therefore $ Time taken by canon $=\frac{50}{2}$
$t ^{\prime \prime}=25\, s$
$\therefore $ Total time taken by a packet to reach on the ground
$= t ^{\prime \prime}+ t + t ^{\prime} $
$=25+10+10$
$=45 \,s .$