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Q.

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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a

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

b

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

c

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

d

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

answer is A, B, C.

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Detailed Solution

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

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Water jet is projected at an angle to the horizontal. At the point of projection, the area of the jet is S1 and at the highest point, the area of the jet is S2 . The initial velocity of projection is u.