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A particle is projected at an angle of `30^(@)w.r.t.` horizontal with speed `20m//s:(` use `g=10m//s^(2))` `(i)` Find the position vector of the particle after `1s`. `(ii)` Find the angle between velocity vector and position vector at `t=1s`. |
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Answer» Correct Answer - (i) `1-0sqrt(3)hat(i)+5hat(j)` (ii) `cos^(-1)(2sqrt((3)/(13)))` `(i) x=ucos theta t` `=20xx(sqrt(3))/(2)xxt=10sqrt(3)m` `y=u sin theta t -(1)/(2)xx10xxt^(2)` `=20xx(1)/(2)xx(1)-5(1)^(2)=5m` Position vector, `vec(r)=10sqrt(3)hat(i)+5hat(j)` `|vec(r)|=sqrt((10sqrt(3))^(2)+5^(2))` `(ii) v_(x)=10sqrt(3)hat(i)` `v_(y)=u_(y)+a_(y)t=10-g t=0` `:. vec(v)=10sqrt(3)hat(i), |vec(v)|=10sqrt(3)` `vec(v).vec(r)=(10sqrt(3)hat(i)).(10sqrt(3)hat(i)+5hat(j))=300` `vec(v).vec(r)=|vec(v)||vec(r)|cos theta` `rArr cos theta =(vec(v).vec(r))/(|vec(v)||vec(r)|)=(300)/(10sqrt(3)sqrt(325))` `rArr theta =cos ^(-1)(2sqrt((3)/(13)))` |
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