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

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.

$u_y\ge \sqrt{2gh}$

$\ge \sqrt{2\times 10\times 500}\ge 100{ \mathrm{ms}}^{-1}$

But   $u^2=u_x^2+u_y^2$

$\therefore$   Horizontal component of initial velocity

$u_x=\sqrt{u^2-u_y^2}$

$=\sqrt{(125{)}^2-(100{)}^2}$

$=75{ \mathrm{ms}}^{-1}$

Time taken to reach the top of the hill.

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

Time taken to reach the ground from the top of the hill.
 $t^{\text{'}}=t=10 \mathrm{s}$
Horizontal distance travelled in 10 s,
 $x=u_x\times t=75\times 10$
     =750m
$\therefore$ Distance through which canon has to be moved
   = 800 - 750 = 50 m
Speed with which canon can move = 2 ms^-1
$\therefore$ Time taken by canon. ${\ell }^{\text{'}}=\frac{50}{2}$

$\Rightarrow$                   = 25s
$\therefore$    Total time taken by a packet to reach on the ground

$=t^{\text{'}\text{'}}+t+t^{\text{'}}$

= 25 + 10 + 10 = 45 s