The initial speed of a projectile fired from ground is u . At the highest point during its motion, the speed of projectile is 3 2 u . The time of flight of the projectile is:

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Correct answer: B. (u)/(g)

The initial speed of a projectile fired from ground is u . At the highest point during its motion, the speed of

At maximum height, only horizontal velocity of the particle remains. Let the angle of project be $\theta$ . The horizontal velocity of the particle will remain same, therefore

$ucos\theta = \dfrac{\sqrt{3}u}{2} ⇒ \theta = 30 °$

Now, time of flight of projectile is

$T = \dfrac{2usin\theta}{g} = \dfrac{u}{g}$

The initial speed of a projectile fired from ground is u . At the...

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