TITLE: Time of flight of projectile from launch to landing
QUESTION [0 upvotes]: The formula below is the time of flight ( time of whole journey from launch(0,0) to landing (×,y) ) of a projectile whose initial vertical position is above the point of impact.

I am trying to understand how the right side of the equation is derived. For instance, how do I come up with 2gy$_0$ ?
$\frac{d}{v.cos(\theta)}$ = $\frac{v.sin(\theta)+\sqrt{\left(v.sin(\theta)\right)^2 +2gy_0}}{g}$
Where
g = gravitational acceleration
y$_0$ = initial vertical position (h)
d = entire horizontal distance or range of the flight from launch to landing
v = velocity
$\theta$ = initial launch angle
Thanks

REPLY [1 votes]: HINT
Let consider at first the equation of motion in vertical direction that is

$y(t)=h+v_0 \sin\theta \cdot t-\frac12 g t^2$

then by the condition $y(t)=0$ find the time of landing $t_{L}$.
Finally use that to find x of landing by

$d=x(t_{L})=v_0 \cos \theta \cdot t_{L}$
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