Show that the maximum height and range of a projectile are `(U^(2) sin^(2) theta)/(2g) `and `(U^(2) sin 2 theta )/(g)` respectively where the terms have their ragular meanings .

Explanation

Let a body is projected with an initial velo-city 'u' and with an angle `theta` to the horizontal. Initial velocity along x direction , `u_(x) = u cos theta`
Initial velocity along y direction , `u_(y) = u sin theta`
Horizontal Range : It is the distance covered by projectile along the horizontal between the point fo projection to the point on ground , where the projectile returns again .
It is denoted by R . The horizontal distance covered by the projectile in the to time of flight is called horizontal range.
Therefore , R = u cos `theta xx ` t .
R = u cos `theta xx (2u sin theta)/(g) = (u^(2)(2 sin theta cos theta))/(g)` , But 2 sin `theta cos theta = sin 2 theta`,
`:.` Range (R) ` = (u^(2) sin 2 theta)/(g)`
Angle of projection for maximum range :
For a given velocity of projection , the horizontal range will be maximum , when sin `2theta` = 1 .
`:.` Angle of projection for maximum range is ` 2 theta = 90^(@) or theta = 45^(@)`
`:. R_(max)=(u^(2))/(g)`
Maximum height : The vertical distance covered by the projectile until its vertical component becomes zero .
Applying `v^(2) - u^(2) ` = 2as we have , 0 - `u^(2) sin^(2) theta = 2 ( - g ) H`
`:. ` Maximum height reached , H `=(u^(2) sin^(2) theta)/(2g)`