A particle is projected vertically upwards. Prove that it will be at3//4 of its greatest height at times which are in the ratio 1:3

A particle is projected vertically upwards. Prove that it will be at3/

1.	A particle is projected vertically upwards. Prove that it will be at3//4 of its greatest height at times which are in the ratio 1:3
Answer» Solution :If U is the initial velocity of a PARTICLE while going vertically upwards then maximum height attained , `h=u^(2)/(2g)` If t is the time when particle reaches at a height `(3//4)`h, then using the RELATION `s=ut+1/2at^(2) ` we have `3/4h=1/2ut+(-g)t^(2)` (or)`3/4(u^(2)/(2g))=ut-1/2"gt"^(2)"(or)" t^(2)-(2u)/"gt"+3/4(u^(2))/(g^(2))=0` Solving it for t , we have `t(2u)/g+-sqrt((4U^(2)/g^(2)-4xx1xx3/4u^(2)/g^(2))/2)=u/g+-u/(2g)` Taking negative sign , `t_(1)=u/g-u/(2g)=(3u)/(2g)` Taking positive sign `=t_(2)=u/g+u/(2g)=(3u)/(2g)` `:.t_(1)/t_(2)=((u/2g)/(3u/2g))=1/3`