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]
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]