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`.

Explanation

Given , speed of packets = 125 m/s
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)gesqrt(2gh)`
`gesqrt(2xx10xx500)`
`ge100m//s`
`"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 m//s`
Time taken to reach the top of the hill,
`t=sqrt((2h)/(g))=sqrt((2xx500)/(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)xxt=75xx10=750m`
`:.` Distance through which canon has to be moved = 800-750=50m
Speed with which canon can move =2 m/s
`:." " ` Time taken by canon `=(50)/(2) rArr t''=25 s`
`:.` Total time taken by a packet to reach on the ground `=t'' +t +t'=25 +10 +10 =45 s`