Velocity at maximum height of a projectile is half of its initial velocity u . Its range on horizont

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3-2.Motion in Plane

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The velocity at the maximum height of a projectile is half of its initial velocity $u$. Its range on the horizontal plane is

A

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

B

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

C

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

D

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

Solution

$u\cos \theta = \frac{u}{2}$

$\therefore \,\,\theta = 60^\circ $

Now $R = \frac{{{u^2}\sin (2 \times 60^\circ )}}{g}$

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

Standard 11

Physics

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