Question

A hill is $500m$ high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of $125m{s}^{-1}$ over the hill. The cannon is located at a distance of $800m$ from the foot of the hill and can be moved on the ground at a speed of $2m{s}^{-1}$ 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 $g=10m{s}^{-2}$ )

Answer 1

Answer: 45

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