A projectile is projected along the direction perpendicular to line of greatest slope on the surface of a smooth inclined plane of inclination 30^o. The horizontal range is 120 m. If angle of projection is be 60°, the speed of projection is

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26.3 m/s

52.6m/s

100m/s

10m/s

answer is A.

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Detailed Solution

The component of acceleration due to gravity along greatest

slope is $g^{'} = gsin 30^{\circ} = 10 \times \frac{1}{2} = 5 {ms}^{- 2}$

$\begin{array}{l} ∴ R = \frac{u^{2} \sin 2 α}{g'} or 120 = \frac{u^{2} \sin 120^{\circ}}{5} \\ \Rightarrow u^{2} = \frac{120 \times 5 \times 2}{\sqrt{3}} \\ ∴ u = \sqrt{\frac{1200}{1 .732}} = 26 .3 {ms}^{- 1} \end{array}$

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