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The initial speed of a projectile fired from ground is $\mathrm{u}$ At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2}\mathrm{u}$ . The time of flight of the projectile is :
Option: 1
$\mathrm{\frac{2u}{g}}$

Option: 2
$\mathrm{\frac{u }{2g}}$

Option: 3
$\mathrm{\frac{\sqrt{3}u }{g}}$

Option: 4
$\mathrm{\frac{u }{g}}$

Answers (1)

best_answer

At the highest point -

$\mathrm{u\cos \theta =\frac{\sqrt{3}u}{2}}$

$\mathrm{\theta=30}$

$\mathrm{T=\frac{2u\sin \theta}{g}=\frac{u}{g}}$

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Gaurav

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