Finding the wet area from a water sprinkler with unknown projection angle

[AR1399]

Homework Statement

A water sprinkler sprinkles water all around it. If the velocity of water coming out is v, find the area around the sprinkler getting wet.[/B]

Homework Equations

In projectile motion,
Range = u^2 sin2(theta) / g

The Attempt at a Solution

What I did was to assume each droplet as a projectile with initial velocity v. Looking at it that way, the range of the droplet would be the radius of the circle of area getting wet.
Everything is fine, except that nothing is mentioned about angle of projection of droplets.
I assumed 45 degrees and got the right ans. in objective type qn, but if the same comes for subjective explained ans, then which angle do we assume?
With 45 degrees, my answer was -
Range = v^2 sin90 / g = v^2 / g
So area = pi (v^2 / g)^2 = pi (v^4 / g^2)[/B]

 

[Qwertywerty]

I think the original question should read - " What is the maximum possible area covered by the water ? " .

Considering that , yes , angle taken would be 45° .

 

[PeroK]

You can't assume any angle. The angle must have been omitted from the question by mistake. Alternatively, the question could have asked for the maximum possible area, which then implies 45°.

 

[AR1399]

Hmmm...must be a qn error as u say. Anyway, thnx a lot for the help!

 

[jbriggs444]

You can't assume any angle. The angle must have been omitted from the question by mistake. Alternatively, the question could have asked for the maximum possible area, which then implies 45°.

As I read the intended meaning of the problem, the sprinkler head is spraying randomly at all angles -- a more or less spherical spray. This is weakly implied by the "all around it" clause.