A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v , then what will be the total area around the fountain that gets wet ?

Explanation

Let the angle with the horizontal be `theta` and the horizontal range of water coming out of the fountain be R.
`therefore R=(v^2sin^2theta)/(g)`
The value of R will be maximum when `sin 2 theta =1 `
i.e., `2theta =90^@ or, theta =45^@`[Fig.2.81]

`therefore R_(max) =(v^2)/(g)`
So a circular area of radius `R_(max)` around the fountain gets wet.
`therefore` Total area,
`A=piR_(max)^(2)=pi((v^2)/(g))^2 =(piv^4)/(g^2)`