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A water fountain on the ground sprinkles...

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:

A

`pi(v^(4))/(g^(2))`

B

`(pi)/(2)(v^(4))/(g^(2))`

C

`pi(v^(2))/(g^(2))`

D

`pi(v^(4))/(g)`

Text Solution

Verified by Experts

The correct Answer is:
A
(a):
Maximum range of water at `theta=45^(@)`
`R_("max")=(v^(2))/(g) sin 2(45^(@))=(v^(2))/(g)`
Area covered by fountain
`A=pi (R_("max")^(2))`
`=(pi v^(4))/(g^(2))`
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