There are two values of time for which a projectile is at the same height. The sum of these two times is equal to (T = time of flight of the projectile)

Explanation

Let u be the speed of projection and 0 be the angle of projection with the horizontal
Here, `u_(x) =u cos 0 and u_(y) =u sin 0`

Let the projectile be at point (x,y) at any time t, then horizontal distance traveled by it
`x = u cos 0 xx t`
and vertical distance traveled by it
`y=u sin 0 xx t =(1)/(2)" gt"^(2)`
if h be the height of point P, then for y=h We have `h=u sin 0 xx t =(1)/(2)" gt"^(2)`
or `(1)/(2)" gt"^(2) -u sin 0 xx t +h=0`
or `t^(2) -(2u sin 0)/(g) t +(2h)/(g) =0`
This equation is quadratic in t and has two roots `t_(1) and t_(2)`. Thus there are two values of time for which the height of the projectile is same during the flight of projectile.
`? t_(1) =(u sin 0)/(g) + sqrt((u^(2) sin^(2)0)/(g^(2))-(2h)/(g))`
and `t_(2) =( u sin 0)/(g) -sqrt((u^(2) sin^(2) 0)/(g^(2))-(2h)/(g))`
Here, `t_(1)` corresponds to point `P and t_(2)` corresponds to point 0 in figure
Now, `t_(1)+t_(2)=(2u sin 0)/(g)`= Time of fight.