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

Explanation

Let us consider a projectile, that is, thrown horizontal with an initial velocity `vec(u)` from the top of a tower of height h in the figure .
Motion along horizontal direction : The particle has zero acceleration along x direction. So, the initial velocity `u(x)` remains constant throughout the motion. The distance traveled by the projectile at a time t is given by `x = u_(x) t + (1)/(2) at^(2)`
Since a = 0 along x direction, we get
`x =u_(x)t " ". . . (1)`
Motion along downward direction : Here `u_(y) = 0 ` (initial velocity has no downward component), a = g .The distance y at time y is given by
`y = u_(y) t + (1)/(2) at ^(2)` , we get
`y = (1)/(2) g t^(2) ` [u = 0, a = g] . . . (2)
Substituting the value of t from equation (1) in equation (2) we have
`y = (1)/(2) g""(x^(2))/(u_(x)^(2))= ((g)/(2u_(x)^(2)))x^(2)`
`y = Kx^(2) " " . . . (3)`
Where `K = (g)/(2u_(x)^(2))` is a constant of proportionality .
Equation (3) gives the trajectory of the projectile. Thus , the path followed by the projectile is a parabola .
Horizontal range : The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range. For horizontal motion, we have
`s_(x) = u_(x) t + (1)/(2) at^(2)`
Here, `s_(x) = R` (range), `u_(x) = u, a = 0 ` (no horizontal acceleration) T is time of flight . Then horizontal range = u T.
Since the time of flight T ` = sqrt((2h)/(g))` , we substitute this and we get the horizontal range of the particle as `R = u sqrt((2h)/(g))` .