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

`vecV= 6hati+2hatj" " h =0.4m`
Given `V_(x)=6=u_(x)` (in projectile)
`V_(y)=2=sqrt(u_(y)^(2)-2gh)`
`therefore U_(y)^(2)=4+2gh=4+2(10)/(0.4)=12`
`therefore U_(y) =sqrt(12) ms^(-1)`
angle of projection ` tan theta=(U_(y))/(U_(x))=(sqrt(12))/(3)=(1)/(sqrt(3))`
`therefore theta= 30^(@)`