InterviewSolution
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InterviewSolution
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A body has maximum range `R_1` when projected up the plane. The same body when projected down the inclined plane, it has maximum range `R_2`. Find the maximum horizontal range. Assume equal speed of projection in each case and the body is projected onto the inclined plane in the line of the greatest slope. |
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Answer» Let `theta_0` be the angle of inclined plane with horizontal. Then for upward projection, `R_(max) = (u^2)/(g(1 + sin theta_0)) = R_1` …(i) For downward projection, `R_(max) = (u^2)/(g(1 - sin theta_0)) = R_2` …(ii) For a projection on horizontal surface, we have `R_(max) = (u^2)/(g) = R`(say) ....(iii) To establish a relation between `R,R_1`, and `R_2`, we need to eliminate `theta_0`. From (i) and (ii), we get `(2)/(R) = (1)/(R_1) + (1)/(R_2) rArr R = (2 R_1 R_2)/(R_1 + R_2)`. |
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