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]

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 $$125$$ $$ms-1$$ 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]

Infinity Learn · Accepted Answer

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

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