InterviewSolution
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InterviewSolution
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The equations of motion of a projectile are given by `x=36tm and 2y =96t-9.8t^(2)m`. The angle of projection isA. `sin^(-1)((4)/(5))`B. `sin^(-1)((3)/(5))`C. `sin^(-1)((4)/(3))`D. `sin^(-1)((3)/(4))` |
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Answer» Correct Answer - A Given : `x = 36t, 2y = 96t - 9.8 t^(2)` or `y = 48t - 4.9t^(2)` Let the initial velocity of projectile be u and angle of projection is `theta`. Then, Initial horizontal component of velocity, `u_(x) = u cos theta = ((dx)/(dt))_(t=0) = 36 or u cos theta = 36 " "...(i)` Initial vertical component of velocity, `u_(y) = u sin theta = ((dy)/(dt))_(t = 0) = 48 or u sin theta = 48 " "...(ii)` Dividing (ii) by (i), we get ` tan theta = (48)/(36) = (4)/(3) therefore sin theta = (4)/(5) or theta = sin^(-1) ((4)/(5))` |
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