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

Options

A. `pi v^(4)/g^(2)`
B. `pi/2 v^(4)/g^(2)`
C. `pi v^(2)/g^(2)`
D. `pi v^(4)/g`

Answer

A

Explanation

Maximum horizontal range `R_(max)=v^(2)/g`
`:.` Total area around the fountain that gets wet is
`piR_(max)^(2)`
`pi(v^(2)/g)^(2)=piv^(4)/g^(2)`