For a projectile if the time of flight is `T`, the maximum height is `H` and the horizontal range is `R`. Find the (a) the maximum height in terms of `T` and (b) the maximum horizontal range in terms of `R` and `H`.

Explanation

(a) `T = (2 u sin theta)/(g)` (i)
`H_(max) = H = (u^(2) sin^(2) theta)/(2g)` (ii)
`(H)/(T^(2)) = ((u^(2) sin^(2) theta)/(2g))/((4u^(2) sin^(2) theta)/(g^(2))) = (1)/(8)g`
`H = (1)/(8)g T^(2)`
OR
Coansider the motion in vertical direction:
At the highest point, `v_(y) = 0`.
`y = (1)/(2)at^(2)`
`t = (T)/(2), y = H`
`H = (1)/(2)g((T^(2))/(2)) = (1)/(8)gT(2)`
Recall: if a particle is moving in a straight line under constant acceleration//retardation and the velocity at one end (inintial/final) is zero, then the displacement in time `t` is `s = (1)/(2)at^(2)`.
(b) `R = (u^(2))/(g)(2 sin theta cos theta)`
`R_(max) = (u^(2))/(g)`
We have to eliminate `theta`
`(R^(2))/(H) = (((u^(2))/(g)) (4 sin^(2) theta cos^(2) theta))/((u^(2) sin^(2) theta)/(2g)) = (8 u^(2) cos^(2) theta)/(g)`
`(R^(2))/(8 H) = (u^(2) (1 - sin^(2) theta))/(g) = (u^(2))/(g) - (u^(2) sin^(2) theta)/(g)`
`(R^(2))/(8 H) = R_(max)-2(u^(2) sin^(2) theta)/(2g)`
`=R_(max) - 2H`
`R_(max) = (R^(2))/(8H) + 2H`