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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TestHub · Accepted Answer

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

$u \cos θ = \frac{\sqrt{3} u}{2} \Rightarrow θ = 30 °$

Now, time of flight of projectile is

$T = \frac{2 u \sin θ}{g} = \frac{u}{g}$

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