Define time of flight.

Explanation

(i) Consider an object is thrown horizontally with initial velocity u along x - direction.

(ii) Since acceleration due to gravity acts vertically downwards, velocity along the horizontal x - direction `u_(x)` doesn't change through the motion. Whereas velocity along the y-direction uy is changed.
The path of the projectile :
(a) Motion along x - direction :
(i) The horizontal distance travelled by the projectile at a point P after a time t can be written as,
`s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)`
(ii) Here, `s_(x)=x, u_(x)=u and a_(x)=0,`
Therefore,
`x=ut`
`t=(x)/(u)" ...(1)"`
(b) Motion along y-direction:
(i) The downward distance travelled by the projectile at a point P after a time t can be written as,
`s_(y)=u_(y)+(1)/(2)a_(y)t^(2)`
(ii) Here, `s_(y)=y, u_(y)=0 and a_(y)=g`
Therefore, `y=(1)/(2)g t^(2)`
(iii) Substituting equation (1), we get,
`y=(1)/(2)g((x)/(u))^(2)=((g)/(2u^(2)))x^(2)`
`y=Kx^(2)" ...(2)"`
Where `K=(g)/(2u^(2))` is a constant.
(iv) The equation (2) represents the equation of a parabola. Thus, the path travelled by the projectile is a parabola.
(c) Time of flight `(T_(f))`
(i) The time of flight `(T_(f))` is hte time taken by the projectile to hit the ground after thrown.
(ii) The downward distance travelled by the projectile at a time t can be written as,
`s_(y)=u_(t)t+(1)/(2)a_(y)t^(2)`
(iii) Here substituting the values `S_(y)=h, t=T_(f), u_(y)=0 and a_(y)=g` we get,
`h=(1)/(2)gT_(f)^(2)`
Therefore, `T_(f)=sqrt((2h)/(g))`
(d) Horizontal range : (R)
(i) The horizontal range(R) is the maximum horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground.
(ii) The horizontal distance travelled by the projectile at a time t can be written as,
`s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)`
(iii) Here, `S_(x)=R, u_(x)=u, a_(x)=0 and T_(f)`
`R=uT_(f)`
(iv) Therefore, `R=usqrt((2h)/(g))" "[because T_(f)=sqrt((2h)/(g))]`