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A particle is projected vertically upwar...

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`.

Text Solution

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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`
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