A particle is projected from gound At a height of 0.4 m from the ground, the velocity of a projective in vector form is `vecv=(6hati+2hatj)m//s` (the x-axis is horizontal and y-axis is vertically upwards). The angle of projection is `(g=10m//s^(2))`

Options

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

Answer

C

Explanation

Let u be the initial speed of the particle
Then `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" "(v_(x)=u_(x))`
or `u_(y)^(2)=v_(y)^(2)+2gh`
or `u_(y)^(2)=(2)^(2)+(2)(10)(0.4)=12`
`therforeu_(y)=sqrt(12)` or `2sqrt(3)m//s`
and `u_(x)=v_(y)=6m//s`
and `u_(x)=v_(y)=6m//s`
`thereforetheta=(u_(y))/(u_(x))=(2sqrt(3))/(6)=(1)/(sqrt(3))`
or `theta=30^(@)`.