Water jet is projected at an angle to the horizontal. At the point of projection, the area of the jet is $S_{1}$ and at the highest point, the area of the jet is $S_{2}$ . The initial velocity of projection is $u$ .

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The maximum height reached from the ground is $\frac{u^{2}}{2 g} (1 - \frac{{S^{2}}_{1}}{{S^{2}}_{2}})$

The range on the level ground is $\frac{2 u^{2}}{g} \frac{S_{1}}{S_{2}} \sqrt{1 - \frac{{S^{2}}_{1}}{{S^{2}}_{2}}}$

The angle of projection is $\cos^{- 1} (\frac{S_{1}}{S_{2}})$

The rate of volume flow is $S_{2} u$

answer is A, B, C.

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Detailed Solution

$S_{1} \times u = S_{2} \times u \cos θ \cos θ = \frac{S_{1}}{S_{2}} R = \frac{2 u^{2} \sin θ \cos θ}{g} = \frac{2 u^{2}}{g} \times \frac{S_{1}}{S_{2}} \times \sqrt{1 - \frac{{S^{2}}_{1}}{{S^{2}}_{2}}} h_{m a x} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{u^{2}}{2 g} \times (1 - \frac{{S^{2}}_{1}}{{S^{2}}_{2}})$