InterviewSolution
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A particle is projected in `x-y` plane with `y-`axis along vertical, the point of projection being origin. The equation of projectile is `y = sqrt(3) x - (gx^(2))/(2)`. The angle of projectile is ……………..and initial velocity si ………………… . |
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Answer» Correct Answer - `60,2 m//sec` `y = sqrt(3) x - (gx^(2))/(2)` comparing with `y = tan theta.x - (1)/(2) (gx^(2))/(u^(2)cos^(2) theta)` `tan theta = sqrt(3) rArr theta = 60^(@)` `u^(2) cos^(2) theta = 1 rArr u^(2) xx (1)/(4) = 1` `u = 2m//s` |
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