A hill is ` 500 m` high. Supplies are to be 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 to hill and can be veoved 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 inwhich a pachet can reach on the ground across the hill ? Taje ` g= 10 m//s^2`.

Options

A. 10s
B. 25s
C. 35s
D. 45s

Answer

D

Explanation

Given, speed of packets `= 125 ms^(-1)`
Height of the hill `= 500 m`.
To cross the hill, the vertical component of the velocity should be sufficient to cross such height,
`u_(y) ge sqrt(2gh)`
`ge sqrt(2 xx 10 xx 500) ge 100 ms^(-1)`
But `u^(2) = u_(x)^(2) + u_(y)^(2)`
`:.` Horizontal component of initial velocity,
`u_(x) = sqrt(u^(2) - u_(y)^(2)) = sqrt((125)^(2) -(100)^(2))`
`= 75 ms^(-1)`
Time taken to reach the top of the hill,
`t = sqrt((2h)/(g)) = sqrt((2 xx 500)/(10)) = 10s`
Time taken to reach the ground from the top of the hill
`t' = t = 10s`
Horizontal distance travelled in 10s
`x = u_(x) xx t = 75 xx 10`
`= 750 m`
`:.` Distance through which cannot has to be moved
`= 800 - 750 = 50m`
Speed with which cannot can move `= 2ms^(-1)`
`:.` Time taken by cannot `= (50)/(2)`
`rArr t'' = 25s`
`:.` Total time taken by a packet to reach on the ground
`= t'' +t +t' = 25 +10 +10 = 45s`