Question

a water fountain on the ground sprinkles water all around it. if the speed of water coming out of the fountain is the total area around the fountain that gets wet is: (iitjeemains, 2011)

Answer 1

This question is the application of oblique projectile.
For finding area around the fountain that gets wet , we have require distance between fountain to outermost point that gets wet by fountain.
e.g., we have to find radius or range of that .

We know, range = u²sin2Ф/g [ from oblique projectile formula]
here, u = v , for maximum range Ф = 45°
so, R = v²sin²2(45°)/g = v²/g

Now, area around the fountain that gets wet = πR²
= π(v²/g)² = πv⁴/g²
hope that help you.....

Answer 2

Hey mate ^_^

Rmax = (v^2/g) sin2θ = v^2/g

Area = πR^2 = πv^4/g^2

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Answer: Area = πR^2 = πv^4/g^2
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