We have a hose pipe which disposes water at the speed of `1- ms^(-1)`. The safe distance from a building on fire, on ground is 5m. How high can this water go ? (take : `g=10 ms^(-2)`)

Explanation

Here we must understand that taking ramge of projectile as 10 m and making projectile hit the building when it is at maximum height is wrong. By doing this we are not achiving maximum y for given x=5m. This just makes highest pt. of path to like on x=5, But there may be other path for which y will be maximum for given x. This problem will be solved by using equation of trajectory by putting x= 5 m and maximising y by varying `theta`.
`y=x tan theta - (gx^(2))/(2u^(2) cos^(2) theta)`
Putting we get `x = 5 m`
`y=5 tan theta -(10xx25 sec^(2) theta)/(2xx100)`
`5 tan^(2) theta-20 tan theta+(4y+5)=0`
for real roots discriminant must be positive.
`400-4 xx 5(4y+5) gt 0`
Solving `3.75 ge y`
hence maximum `y=3.75 m`
If we have taken range as 10 m then angle of projection will be `theta=45^(@)` corsponding maximum hight `H=2.5 m` which is smaller than our answer.