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 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.

Answer 1

According to the problem, the speed of packets = 125 m/s, the height of the hill = 500 m, the distance between the cannon and the foot of the hill, d = 800 m

To cross the hill in the shortest time, then the vertical component of the velocity should be minimum so that it just crosses the height of hill.

Distance through which canon has to be moved = 800 – 750 = 50 m

The speed with which canon can move = 2 m/s

`u_y = sqrt(2gh) ≥ sqrt(2 xx 10 xx 500) ≥ 100` m/s

But `u^2 = u_x^2 + u_y^2`

∴ The horizontal component of initial velocity,

`u_x = sqrt(u^2 - u_y^2)`

= `sqrt((125)^2 - (100)^2`

= 75 m/s

Time is taken by the packet to reach the top of the hill,

`t = sqrt((2h)/g)`

= `sqrt((2 xx 500)/10`

= 10 s

Time taken to reach the ground from the top of the hill t' = t = 10 s.

Horizontal distance travelled in 10 s,

`x = u_x xx t`

= `75 xx 10`

= 750 m

So the distance between the canon and the hill is 750 m

Distance for which the canon needs to move = 800 – 750 = 50 m

Time is taken for the canon to move 50 m = `50/2` = 25 sec

So the total time taken by the packet = 25 + 10 + 10 = 45 seconds.