An aeroplane is flying horizontally at a height of `2/3` km with a velocity of 500 km/h. Find the rate at which it is receding from a fixed point on the ground which it passed over 2 min ago.

Explanation

We first find the area in which the cannon shell can reach. The equation of trajectory for cannon shell is
`y=x tantheta-((1)/(2)gx^(2))/(u^(2))sec^(2)theta`.......(`1`)
For maximum `y` for a given value of `x`-
`(dy)/(d(tantheta))=x-((1)/(2)gx^(2))/(u^(2))(2tantheta)=0 rArr tantheta=(u^(2))/(gx)`
Putting in equation (`1`)
`y_(max)=x(u^(2))/(gx)-((1)/(2)gx^(2))/(u^(2))[1+(u^(4))/(g^(2)x^(2))]=(u^(2))/(2g)-((1)/(2)gx^(2))/(u^(2))`
`:. ` The cannon shell can hit an area given by
`y le (u^(2))/(2g)-((1)/(2)gx^(2))/(u^(2))`
Given in the problem `y=250m`, `u=100m//s`, `g=10m//s`.
Putting these value we get,
`(x^(2))/(2000) le 250 rArr -500sqrt(2) le x le 500sqrt(2)`
`:.` Plane is in danger for a period of `(1000sqrt(2))/(500)=2sqrt(2)sec`