A projectile is projected from horizontal with velocity (u) making an angle ` 45^(@)` with the horizontal direction. Find the distance of the highest point of the projectile from its starting point.

Explanation

Maximum height ,
` H= (u^2 sin ^2 45^@)/(2g) = u^2/(2 g) ( 1 /(sqrt2) ^(2) = u^2/( 4g)`
Horizontal range, ` R = (u^2 sin 2 xx 45^2)/g = u^2/g`
In Fig. 2 (d) . 36 , AC =H =u^2 /(4g)`
. ltvegt ` OC =R/2 = u^2/ (2 g)`
` OA =sqrt (OC^2 + A C^2 ) = sqrt ((u^2/(2 g) )^(2) + (u^2/(4 g))^2)`
`= u^2/(2 g) sqrt 1 + 1/4 = (sqrt 5 )/4 u^2/g`.