A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountains 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^(2))/(g)`

Answer

A

Explanation

Total area around fountain
` A = pi R_(max)^(2) = pi (v^(4))/(g^(2)) `
` [ because R_(max) = (v^(2) sin 2theta )/(g) = (v^(2) sin 2 theta)/ (g) = (v^(2) sin 90(@))/ (g) = (v^(2))/(g) ]`