A particle is moving along a vertical circle of radius $R = 20 m$ with a constant speed $v = 31.4 {ms}^{- 1}$ as shown in Fig. Straight line ABC is horizontal and passes through the centre of the circle. A shell is fired from point A at the instant when particle is at C.
If distance AB is $20 \sqrt{3} m$ and shell collides with the particle at B, ( $π = 3.14$ and $g = 10 {m/s}^{2}$ )

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smallest possible value of the angle $θ$ of projection, will be $30^{\circ}$

velocity u of projections should be 20 m/s

smallest value of $θ$ must be $45^{\circ}$

velocity of projectile will be 30 m/s

answer is A, B.

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Detailed Solution

Number of revolutions completed must be odd multiples of half
$T = 4 \sec,$ T. time of light of projectile
$T_{0} = \frac{o d d}{2} T$
$\frac{24 \sin θ}{g} = \frac{(o d d) \times 4}{2}$ (1)
And $\frac{4^{2} \sin 2 θ}{g} = 20 \sqrt{3}$ (2)
Solving (1) and (2) and $θ$ smallest
We get $\tan θ = \frac{1}{\sqrt{3}} θ = 30$
And $4 = 20 m/s.$