A projectile is projected with speed u at an | vedclass

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(D) Let the angle of inclination of the plane be $\alpha$. The velocity of the projectile is $u$ at $60^o$ with the horizontal.When the projectile hit

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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 velocity of the projectile is $u$ at $60^o$ with the horizontal.When the projectile hits the plane horizontally,its vertical component of velocity becomes zero at the point of impact.Let the horizontal and vertical components of velocity be $v_x = u \cos 60^o = u/2$ and $v_y = u \sin 60^o = u\sqrt{3}/2$.At the point of impact,the vertical velocity $v_y' = v_y - gt = 0$,so $t = \frac{u\sqrt{3}}{2g}$.The horizontal distance covered is $x = v_x t = (u/2) 	imes (u\sqrt{3}/2g) = \frac{u^2\sqrt{3}}{4g}$.The vertical height reached is $y = v_y t - \frac{1}{2}gt^2 = \frac{u\sqrt{3}}{2} \cdot \frac{u\sqrt{3}}{2g} - \frac{1}{2}g \left(\frac{u\sqrt{3}}{2g}ight)^2 = \frac{3u^2}{4g} - \frac{3u^2}{8g} = \frac{3u^2}{8g}$.The angle of inclination $\alpha$ satisfies $	an \alpha = y/x = (3u^2/8g) / (u^2\sqrt{3}/4g) = \sqrt{3}/2$.The range on the inclined plane is $R_{inc} = \frac{x}{\cos \alpha}$. Since $	an \alpha = \sqrt{3}/2$,then $\cos \alpha = \frac{2}{\sqrt{7}}$.Thus,$R_{inc} = \frac{u^2\sqrt{3}}{4g} \cdot \frac{\sqrt{7}}{2} = \frac{u^2\sqrt{21}}{8g}$.

$A$ 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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