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A stone is projected from the ground with a velocity of $14 {ms}^{- 1}$ one second later it clears a wall 2m high. The angle of projection is $(g = 10 {ms}^{- 2})$

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15 °

30 °

60 °

45 °

answer is B.

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Detailed Solution

y = u_{y} t - \frac{1}{2} {gt}^{2}

$y = u \sin θ t - \frac{1}{2} {gt}^{2}$

$y = 14 \sin θ (1) - \frac{1}{2} \times 10 \times 1$

$2 + 5 = 14 \sin θ$

$\sin θ = \frac{7}{14} = \frac{1}{2}$

$θ = 30 °$

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