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Let $R_1$ and $R_2$ respectively be the maximum ranges up and down an inclined plane and $R$ be the maximum range on the horizontal plane. Then $\mathrm{R}_1, \mathrm{R}, \mathrm{R}_2$ are in
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Н.P
Н.P
Let $\beta$ be the inclination of the plane to the horizontal and $u$ be the velocity of projection of the projectile
$\mathrm{R}_1=\frac{\mathrm{u}^2}{\mathrm{~g}(1+\sin \beta)}$ and $\mathrm{R}_2=\frac{\mathrm{u}^2}{\mathrm{~g}(1-\sin \beta)}$
$\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2 \mathrm{~g}}{\mathrm{u}^2}$ or $\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}\left[\because \mathrm{R}=\frac{\mathrm{u}^2}{\mathrm{~g}}\right]$
Therefore, $\mathrm{R}_1, \mathrm{R}, \mathrm{R}_2$ are in H.P.
$\mathrm{R}_1=\frac{\mathrm{u}^2}{\mathrm{~g}(1+\sin \beta)}$ and $\mathrm{R}_2=\frac{\mathrm{u}^2}{\mathrm{~g}(1-\sin \beta)}$
$\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2 \mathrm{~g}}{\mathrm{u}^2}$ or $\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}=\frac{2}{\mathrm{R}}\left[\because \mathrm{R}=\frac{\mathrm{u}^2}{\mathrm{~g}}\right]$
Therefore, $\mathrm{R}_1, \mathrm{R}, \mathrm{R}_2$ are in H.P.
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