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A particle projected from O and moving freely under gravity strikes the horizontal plane passing through O at a distance R, from starting point O as shown in the figure. Then:

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there will be two angles of projection if $Rg < u^{2}$

there will be more than two angles of projection if $Rg < u^{2}$

the two possible angles of projection are complementary

the product of the possible times of flight from O to A is 2R/g

answer is A, C, D.

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Detailed Solution

$R = \frac{u^{2} \sin 2 θ}{g} or \frac{Rg}{u^{2}} = \sin 2 θ < 1$

Therefore, there shall be two values of $θ$ , that is $θ_{1} and 90^{\circ} - θ_{1}$ for same range.

It is obvious $θ_{1} and 90^{\circ} - θ_{1}$ are complementary.

$\begin{array}{l} T_{1} = \frac{2 usin θ_{1}}{g} and T_{2} = \frac{2 usin (90^{\circ} - θ_{1})}{g} = \frac{2 ucos θ_{1}}{g} \\ \Rightarrow T_{1} T_{2} = \frac{2 u^{2} \sin 2 θ_{1}}{g^{2}} = \frac{2}{g} R \end{array}$

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