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For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then find the angle of inclination of the inclined plane. |
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Answer» Correct Answer - 3 3 `(u^(2))/(g[1-sinbeta])=(3u^(2))/(g[1+sin beta])implies1+sinbeta=3-3sinbetaimpliessinbeta=1/2impliesbeta=30^(@)` |
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