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 range. Assume the equal speed of projection in each case and the body is projected onto the greatest slope.

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$R = \frac{R_{1} R_{2}}{R_{1} - R_{2}}$

$R = \frac{4 R_{1} R_{2}}{R_{1} + R_{2}}$

$R = \frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$

$R = \frac{2 R_{1} R_{2}}{R_{1} - R_{2}}$

answer is B.

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Detailed Solution

For upward projection, $R_{m a x} = \frac{V_{0}^{2}}{g (1 - \sin β)} = R_{1}$ ………..(i)

For downward projection, $R_{m a x} = \frac{v_{0}^{2}}{g (1 + \sin β)} = R_{2}$ ……….(ii)

For a projection of horizontal surface substituting $β = 0$ , Then, we have, $R_{m a x} = \frac{V_{0}^{2}}{g} = R (s a y)$ ……..(iii)

To establish a relation between $R, R_{1}$ and $R_{2}$ we need to eliminate $\sin β$ .

Adding $\frac{1}{R_{2}}$ form eq.(i) with $\frac{1}{R_{2}}$ from eq.(ii) we have $\frac{2}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ Then,

Ans- $R = \frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$

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