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`.

Answer

(i) `1-0sqrt(3)hat(i)+5hat(j)`
(ii) `cos^(-1)(2sqrt((3)/(13)))`

Explanation

`(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)))`