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    				| 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)` | |