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(D) Let the angle of inclination of the plane be $\alpha$. The projectile is fired at $	heta = 60^o$ with the horizontal. Since it hits the plane hor

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$\frac{\sqrt{21}}{8} \frac{u^2}{g}$

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(D) Let the angle of inclination of the plane be $\alpha$. The projectile is fired at $	heta = 60^o$ with the horizontal. Since it hits the plane horizontally,the vertical component of velocity at the point of impact is $0$. Thus,$v_y = u \sin 	heta - gt = 0$,which gives $t = \frac{u \sin 60^o}{g} = \frac{u \sqrt{3}}{2g}$.At this time $t$,the horizontal displacement is $x = u \cos 60^o \cdot t = \frac{u}{2} \cdot \frac{u \sqrt{3}}{2g} = \frac{u^2 \sqrt{3}}{4g}$.The vertical displacement is $y = u \sin 60^o \cdot t - \frac{1}{2}gt^2 = \frac{u \sqrt{3}}{2} \cdot \frac{u \sqrt{3}}{2g} - \frac{1}{2}g \left(\frac{u \sqrt{3}}{2g}ight)^2 = \frac{3u^2}{4g} - \frac{3u^2}{8g} = \frac{3u^2}{8g}$.The inclination angle $\alpha$ satisfies $	an \alpha = \frac{y}{x} = \frac{3u^2/8g}{u^2\sqrt{3}/4g} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.The range $R$ on the inclined plane is given by $R = \frac{x}{\cos \alpha}$. Since $	an \alpha = \frac{\sqrt{3}}{2}$,$\cos \alpha = \frac{2}{\sqrt{2^2 + (\sqrt{3})^2}} = \frac{2}{\sqrt{7}}$.$R = \frac{u^2 \sqrt{3}/4g}{2/\sqrt{7}} = \frac{u^2 \sqrt{21}}{8g}$.

$A$ projectile is projected with speed $u$ at an angle of $60^o$ with the horizontal from the foot of an inclined plane. If the projectile hits the inclined plane horizontally,the range on the inclined plane will be:

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