1.

A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

Answer» Correct Answer - A::C
Given, `v_|_u`
`rArr v.u=0`
`rArr (u+at).u=0`......(i)

Substituting the proper values in Eq.(i), we have
`[{( ucos theta)hat(i)(usin theta)hat(j)}+(-ghat(j))t].[(u cos theta)hat(i)+(u sin theta)hat(j)]=0`
`rArr u^(2)cos^(2)theta+u^(2)sin^(2)theta-(ug sin theta)t=0`
`rArr u^(2)(sin^(2) theta+cos^(2) theta)=(ug sin theta)t`
Solving this equation, we get
`t=(u)/(g sin theta)=(u cosec theta)/(g)`


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