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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)` |
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