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(D) Initial velocity of the projectile, $u = 10 \,m/s$. The angle of projection with the horizontal is $60^{\circ}$ and the inclination of the plane i

Vedclass · Accepted Answer

$\frac{20}{3} \,m$

Vedclass · Answer

(D) Initial velocity of the projectile, $u = 10 \,m/s$. The angle of projection with the horizontal is $60^{\circ}$ and the inclination of the plane is $\alpha = 30^{\circ}$. The angle of projection relative to the inclined plane is $	heta = 60^{\circ} - 30^{\circ} = 30^{\circ}$. The formula for the range $R$ along an inclined plane is: $R = \frac{2 u^2 \cos 	heta \sin(	heta - \alpha)}{g \cos^2 \alpha}$ Wait, the standard formula for range up an incline is $R = \frac{2 u^2 \cos 	heta \sin(	heta - \alpha)}{g \cos^2 \alpha}$. Using $	heta = 60^{\circ}$ (angle with horizontal) and $\alpha = 30^{\circ}$ (angle of incline): $R = \frac{2 u^2 \cos 60^{\circ} \sin(60^{\circ} - 30^{\circ})}{g \cos^2 30^{\circ}}$ $R = \frac{2 	imes (10)^2 	imes (1/2) 	imes \sin 30^{\circ}}{10 	imes (\sqrt{3}/2)^2}$ $R = \frac{200 	imes 0.5 	imes 0.5}{10 	imes 0.75} = \frac{50}{7.5} = \frac{500}{75} = \frac{20}{3} \,m$.

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