A projectile is fired horizontally with a velocity u making an angle `theta` . Derive the expression for time of the flight .

Explanation

Time of flight is the total taken by the projectile from the point of projection till it hits the horizontal plane.
This time of flight is given by the time taken by the projectile to go from point O to B via point A in the figure .Here initial velocity is resolved into components as follows .
We know that
`s_(y) = u_(y) t = (1)/(2) a_(y) t^(2)`
Here, `s_(y) = y = 0` (net displacement in y-direction is zero),
`u_(y) = u sin theta, a_(y) = - g, t = T_(f)`
Then `0 = u sin theta T_(f) -(1)/(2) g T_(f)^(2)`
`T_(f) = 2u (sin theta)/(g)`