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Q.

The speed at the maximum height of a projectile is half of its initial speed of projection (u). The horizontal range of the projectile is

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a

$\frac{\sqrt{3} u^{2}}{2 g}$

b

$\frac{\sqrt{3} u^{2}}{g}$

c

$\frac{u^{2}}{\sqrt{3} g}$

d

$\frac{2 u^{2}}{\sqrt{3} g}$

answer is A.

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Detailed Solution

$minimum velocity ucosθ = \frac{u}{2} θ = 60^{\circ} ∴ R = \frac{u^{2} \sin 2 θ}{g} = \frac{\sqrt{3} u^{2}}{2 g}$

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