At a height `0.4 m` from the ground the velocity of a projectile in vector form is `vec v = (6 hat i+ 2 hat j) ms ^-1`. The angle of projection is

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. `tan^(-1)(3/4)`

Answer

c

Explanation

`v^(2)=u^(2)-2gh or u^(2)=v^(2)+2gh`
or `u_(x)^(2)+u_(y)^(2)=v_(x)^(2)+v_(y)^(2)+2gh`
As `v_(x)=u_(x), u_(y)^(2)=v_(y)^(2)+2gh`
or `u_(y)^(2)=(2)^(2)+2xx10xx0.4=12`
`u_(y)=sqrt(12)=2sqrt(3)m//s`
`u_(x)=v_(x)=6m//s`
`tan theta=(u_(y))/(u_(x))=(2sqrt(3))/6=1/(sqrt(3))`
`theta=30^(@)`