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(D) The water fountain acts as a projectile source spraying water in all directions with speed $v$. The maximum horizontal range $R_{\max}$ of the wat

Vedclass · Accepted Answer

$\frac{\pi v^4}{g^2}$

Vedclass · Answer

(D) The water fountain acts as a projectile source spraying water in all directions with speed $v$. The maximum horizontal range $R_{\max}$ of the water is achieved when the angle of projection is $45^\circ$. Using the formula for horizontal range: $R = \frac{v^2 \sin(2	heta)}{g}$. For maximum range,$\sin(2	heta) = 1$,so $R_{\max} = \frac{v^2}{g}$. The water covers a circular area around the fountain with radius $R_{\max}$. Therefore,the total area $A$ is given by: $A = \pi R_{\max}^2 = \pi \left( \frac{v^2}{g} ight)^2 = \frac{\pi v^4}{g^2}$.

$A$ water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$,the total area around the fountain that gets wet is :

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