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A body has maximum range R_(1) when projected up the inclined plane. The same body when projected down the inclined plane, it has maximum range R_(2). Find its maximum horizontal inclined plane in the line of the greatest slope. |
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Answer» Solution :As derived earlier, For upward projection, `R_("max") =(v_(0)^(2))/(g(1+sinbeta))= R_(1)""....(i)` For downward projection, `R_("max") =(v_(0)^(2))/(g(1-sin beta)) = R_(2)""......(iii)` For a projection on horizontal surface substituting `beta = 0` Then, we have `R_("max") = (v_(0)^(2))/(g) = R("SAY")"".......(iii)` To establish a relation between R, `R_(1)` and `R_(2)` we need to eliminate `sin beta` ADDING `(1)/(R_(1))`from eq. (i) with`(1)/(R_(2))` from eq (ii) we have `(2)/(R) = (1)/(R_(1)) + (1)/(R_(2))` Then `R= (2R_(1)R_(2))/(R_(1)+R_(2))` |
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