A particle is projected in xy plane with y-axis along vertical, the point of projection is origin. The equation of the path is $y = \sqrt{3} x - \frac{g}{2} x^{2}$ . where y and x are in m. Then the speed of projection in ms^-1 is

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$\sqrt{3}$

$\frac{\sqrt{3}}{2}$

answer is A.

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Detailed Solution

Compare the equation with

$y = xtan θ - \frac{g}{2 u^{2} \cos^{2} θ} x^{2}$ here $x \tan θ = \sqrt{3} x, θ = 60^{0} a n d \frac{g}{2} x^{2} = \frac{g}{2 u^{2} \cos^{2} θ} x^{2}$

u= 2m/s

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