Q. 293.8( 4 Votes )

# 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 800m 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/s^{2}.

Answer :

Given:

Height of the hill = h = 500m

Speed of projection of packets = u = 125m/s

Distance of canon = 800m

Speed of canon = v = 2m/s

For time to be minimum, the packet should just pass the top of the hill and so the vertical component of velocity must be minimum.

So,

Now let’s calculate the horizontal component of velocity.

Now, for time to reach the top of the hill,

Hence, 20 seconds is the time of flight.

Now, let’s find the horizontal distance between the canon and top of the hill.

R = u_{x}t = 75m/s×10s =750m

Distance travelled by canon = 800m – 750m = 50m

Time taken by the canon *=*

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PREVIOUSIf |A|= 2 and |B| = 4, then match the relations in column I with the angle θ between A and B in column IINEXTA gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R = .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig 4.5), show that it could be achieved by raising the gun to a height at leasth = ∆x(Hint: This problem can be approached in two different ways:(i) Refer to the diagram: target T is at horizontal distance x = R + ∆x and below point of projection y = – h.(ii) From point P in the diagram: Projection at speed vo at an angle θ below horizontal with height h and horizontal range ∆x.)

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