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.

Options

A. `R=(2R_(1)R_(2))/(R_(1)-R_(2))`
B. `R=(2R_(1)R_(2))/(R_(1)+R_(2))`
C. `R=(R_(1)R_(2))/(R_(1)-R_(2))`
D. `R=(4R_(1)R_(2))/(R_(1)+R_(2))`

Answer

B

Explanation

As derive earlier, for upward projection, `R_(max)=(v_(0)^(2))/g(1-sin beta)=R_(1)`…(i)
For downward projection,
`R_(min)=(v_(0)^(2))/g(1-sin beta)=R_(2)`…(ii)
For a projection of horizontal surface substituting `beta=0`,Then, we have `R_(max)=(v_(0)^(2))/g(1-sin beta)=R(say)`…(iii)
To establish a relation between `R,R_(1)` and `R_(2)` , we need to eliminate `sin beta`.Adding `1/R_(1)` form eq. (i) with `1/R_(2)` from eq.(ii) we have `2/R=1/R_(1)+1/R_(2)` then, ans . `R=(2R_(1)R_(2))/(R_(1)+R_(2))`