Show that the maximum height reached by a projectile launched at an angle of `45^@` is one quarter of its range.

Explanation

In projectiles Range, ( R )= `(u^2sin2theta)/g`
Maximum height (H)= `(u^2sin^2theta)/(2g)`
When `theta=45^@`
`impliesR=R_(max)=(u^2sin90^@)/g`
`R_(max)=u^2/g`
Maximum height
`H_(max)=(u^2sin^2(45^@))/(2g)`
`=(u^2(1//sqrt2)^2)/(2g)=u^2/(4g)`
`H_(max)=1/4R_(max)`
`therefore` When `theta=45^@` maximum height reached is one quarter of maximum range.