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  • 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 over the hill.

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 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 m/s 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 g = 10 m/s2.

Answers (1)

Packet speed = 125 m/s, height of hill = 500m

In order to cross the hill, the vertical component of the packet should be reduced to make the height of 500m attainable. The distance between canon and hill should also be half of that of the packet’s range.

V^{2}-u^{2}=2gh

U_{y}=\sqrt{2gh}=\sqrt{10000}=100\; m/s

Now, U=u{_{x}}^{2}+u{_{y}}^{2}

u{_{x}}^{2}=125\times125-(100\times100)

u_{x}=75\; m/s

now we consider the packet's vertical motion,

v_{y}=u_{y}+gt

t= total time of flight = 10 sec

so, v_{y}=75\times10=750

so the distance between canon and hill is 750m

distance for which the canon needs to move=800-750=50m

time taken for the canon to move 50m=\frac{50}{2}=25 sec

so the total time taken by the packet =25+10+10=45\; seconds

Posted by

infoexpert23

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