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.

Explanation

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))`