Define time of flight.

Explanation

Maximum Height : `h_(max)`
(i) The maximum vertical distance travelled by the projectile during its journey is called maximaum height.
(ii) For the veritcal part of the motion, `v_(y)^(2)= u_(y)^(2)+2a_(y)s_(y)`.
(iii) Hence, `V_(y) =0, S_(y)= h_(max), u_(y) =u " and "a_(y)= -g`
Therefore, `0=u^2 sin^2theta -2gh_(max)`
`h_(max)= (u^2 sin^2 theta)/(2g)`
Time of fligt : `(T_f)`
(i) Time time of flight `(T_f)` is the time taken by the projectile to hit the ground after thrown.
The downward distance travelled by the projectile at a time t can be written as,
`S_(y)=u_(y)t +(1)/(2)a_(y)t^(2)`
(iii) Here substituting the values `S_(y)=0, t=T_(f),u_(y)= u sin theta, " and "a_(y) = -g` we get,
`0= u sin theta - (1)/(2) gT_(f)^(2)`
Therefore, `T_(f) = (2u sin theta)/(g)`
Horizontal range : (R)
(i) The horizontal range (R ) is the maximum horizontal distance btween the point of projectile and 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 cos theta, a_(x)=0 " and "t=T_(f)R= u cos theta*T_(f)`
`R= u cos theta *(2u sin theta)/(g)= (2u^2 sin theta cos theta)/(g)" "[:. T_(f) = (2u sin theta)/(g)]`
Therefore, `R= (u^2 sin 2 theta)/(g)" "[:. sin2 theta = 2sin theta.cos theta]`
(iv) For maximum range, `sin2theta =1`
`2theta= (pi)/(2)`
`theta = (pi)/(4)`
The maximum range is, `R=(u^2)/(g)`.