A water jet directed by firemen must reach a window 40 m above the ground , from a nozzle of diameter 3 cm discharging 30 kg/s .Determine the largest distance from the building where a fireman can stand and still reach the jet into the window .

Explanation

Taking O as the origin of coordinate system . The equation of the water jet is
` y = x tan alpha - (gx^(2))/(2u^(2)) sec^(2) alpha`
Here, y = 38 (constant )
` rArr 38 = x tan alpha - (gx^(2))/(2u^(2)) sec^(2)alpha`
Differentiating with respect to `alpha ` , we get
` 0 = (dx)/(d alpha) tan alpha + x sec^(2) alpha `
Differentiating with respect to `alpha` , we get
` 0 (dx)/(d alpha) tan alpha + x sec^(2)alpha - g/(2u^(2)) { 2x (dx)/(d alpha) sec^(2) alpha +x^(2) 2 sec^(2) alpha . tan alpha}`
Now , for x to be maximum , `(dx)/(d alpha) = 0 `
or `tan alpha = (u^(2))/(gx)`
It can be shown that his value of `alpha` gives the maximum for x
` y = (xu^(2))/(gx) - (gx^(2))/(2u^(2)) (1+u^(4)/(g^(2)x^(2)))`
` rArr x_(max)^(2) = (2u^(2))/g ((u^(2))/(2g) - y)`
Now , `pi/4 (3xx10^(-2))^(2) xx 1 xx 10^(3) xx u = 30 `
`rArr u = (1200)/(9pi) m/s = 42.44 ` m/s
Putting this value of u and y = 38 m ew get `X_(max)= 137 m `