1.

The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]` Angle of projection `theta` is :-A. `30^(@)`B. `60^(@)`C. `45^(@)`D. `sqrt(3)` rad

Answer» Correct Answer - B
`y=sqrt(3)x-2x^(2)`
Trajectory equation is `y=x tan theta-(gx^(2))/(2u^(2) cos^(2) theta)`
`tan theta=sqrt(3)rArr theta=60^(@)` & `(g)/(2u^(2)cos^(2) theta)=2`
`rArr u=5/(2xx1/4)=sqrt(10)`


Discussion

No Comment Found

Related InterviewSolutions