The maximum height reached by a projectile is h. It’s time of flight is
(A) $\sqrt {\dfrac{{4h}}{g}} $
(B) $\dfrac{{8h}}{g}$
(C) $\sqrt {\dfrac{{8h}}{g}} $
(D) $\sqrt {\dfrac{{16h}}{g}} $

Hint First use the formula for the maximum height when a projectile is fired at an angle θ with the horizontal to calculate the value of \[uSin\theta \]which is given by the following relation
$h = \dfrac{{{u^2}{{\operatorname{Sin} }^2}\theta }}{{2g}}$
Where, h = Maximum height reached by the projectile.
     u = Initial velocity of the projectile.
     θ = Angle made by the projectile with the horizontal.
     g = Acceleration due to gravity.
Then use the formula for calculating the time of ascent which is given by the following relation
$t = \dfrac{{2u\operatorname{Sin} \theta }}{g}$
Where, t = time of ascent
u = Initial velocity of the projectile.
     θ = Angle made by the projectile with the horizontal.
     g = Acceleration due to gravity.

Complete step by step solution
We know for a projectile Maximum height is given by
$h = \dfrac{{{u^2}{{\operatorname{Sin} }^2}\theta }}{{2g}}$……(i)
Where, h = Maximum height reached by the projectile.
     u = Initial velocity of the projectile.
     θ = Angle made by the projectile with the horizontal.
     g = Acceleration due to gravity.
From equation (i) on cross multiplying we have
${u^2}{\operatorname{Sin} ^2}\theta = 2gh$
Taking square root on both sides
$u\operatorname{Sin} \theta = \sqrt {2gh} $……(ii)
We know that time of ascent or flight is given by
$t = \dfrac{{2u\operatorname{Sin} \theta }}{g}$……(iii)
Where, t = time of ascent
u = Initial velocity of the projectile.
     θ = Angle made by the projectile with the horizontal.
     g = Acceleration due to gravity.
Substituting the value of \[uSin\theta \]from equation (ii) into equation (iii) we get
$t = \dfrac{2}{g}\sqrt {2gh} $
Moving everything on the RHS inside the square root we get
Or, $t = \sqrt {\dfrac{{8h}}{g}} $. i.e.

Correct option is option C.

Note It should be noted that the initial velocity has two components \[u\operatorname{Cos} \theta \](horizontal component) and \[uSin\theta \](vertical component). However as ‘g’ (acceleration due to gravity) has no horizontal component which makes the horizontal component uniform while the vertical component is non-uniform as ‘g’ acts exactly opposite to it.