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-1Height of the hill = 500 mTo cross the hill, the vertical component of the velocity should be sufficient to cross such height.uy≥2gh≥2×10×500≥100 ms-1But   u2=ux2+uy2∴   Horizontal component of initial velocityux=u2-uy2=(125)2-(100)2=75 ms-1Time taken to reach the top of the hill.t=2hg=2×50010=10 sTime taken to reach the ground from the top of the hill. t'=t=10 sHorizontal distance travelled in 10 s, x=ux×t=75×10     =750m∴ Distance through which canon has to be moved   = 800 - 750 = 50 mSpeed with which canon can move = 2 ms-1∴ Time taken by canon. ℓ'=502⇒                   = 25s∴    Total time taken by a packet to reach on the ground=t''+t+t'= 25 + 10 + 10 = 45 s