If the maximum range down an inclined plane i | vedclass

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Question

(A) The maximum range of a projectile on an inclined plane with angle of inclination $\alpha$ is given by:For upward projection: $(R_{\max})_{up} = \f

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The maximum range of a projectile on an inclined plane with angle of inclination $\alpha$ is given by:For upward projection: $(R_{\max})_{up} = \frac{u^2}{g(1 + \sin \alpha )}$For downward projection: $(R_{\max})_{down} = \frac{u^2}{g(1 - \sin \alpha )}$According to the problem,$(R_{\max})_{down} = 3 	imes (R_{\max})_{up}$.Substituting the expressions: $\frac{u^2}{g(1 - \sin \alpha )} = 3 	imes \frac{u^2}{g(1 + \sin \alpha )}$Simplifying: $1 + \sin \alpha = 3(1 - \sin \alpha )$$1 + \sin \alpha = 3 - 3 \sin \alpha$$4 \sin \alpha = 2$$\sin \alpha = \frac{1}{2}$Therefore,$\alpha = 30^o$.

If the maximum range down an inclined plane is three times the maximum range up the same inclined plane,find the angle of inclination in degrees.

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