A particle is projected vertically upwards. Prove that it will be at three-fourth of its greatest height at times which are in the ratio `1: 3`.

Explanation

If u is the initial velocity of a particle while going vertically upwards .
Then maximum height attained,h`(u^(2))/(2g)`
If t is time when particle reaches at a height (3/4)h, then using the relation
`s=ut+(1)/(2)at^(2)`.
We have `(3)/(4)h=(1)/(2)ut+(-g)t^(2)`
(or) `(3)/(4)(u^(2))/(2g)=ut-(1)/(2)"gt"^(2)` (or) `t^(2)-(2u)/(g)t+(3)/(4)(u^(2))/(g^(2))=0`
Solving it for t,we have
`t=((2u)/(g)pmsqrt((4u^(2))/(g^(2))-4xx1xx(3)/(4)(u^(2))/(g^(2))))/(2)=(u)/(g)pm(u)/(2g)`
Taking negative sign ,`t_(1)=(u)/(g)-(u)/(2g)=(u)/(2g)`
Taking negative sign =`t_(2)=(u)/(g)+(u)/(2g)=(3u)/(2g)`
`therefore (t_(1))/(t_(2)) =((u//2g)/(3u//2g))=(1)/(3)`