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Comparing $y=\sqrt 3 x-\large\frac{1}{2} $$x^2$

with $y=\tan\theta x-\large\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta}$$x^2$

$\tan \theta=\sqrt 3=>\theta=60 ^{\circ}$

and $\large\frac{1}{2}=\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta}$

=>$u^2\cos ^2 \theta=g$

$u^2\cos ^2 60=10$

$u^2.\large\frac{1}{4}=10$

$u^2=40$

$u=2 \sqrt {10}$

Hence b is the correct answer.

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