JEE MAIN - Physics (2011 - No. 26)
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 :
$$\pi {{{v^4}} \over {{g^2}}}$$
$${\pi \over 2}{{{v^4}} \over {{g^2}}}$$
$$\pi {{{v^2}} \over {{g^2}}}$$
$$\pi {{{v^2}} \over g}$$
Explicació
Maximum range of water coming out of fountain,
$${R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}$$
Total area around fountain,
$$A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}$$
$${R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}$$
Total area around fountain,
$$A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}$$