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(D) The range $R$ of a projectile on an inclined plane is given by the formula:$R = \frac{2 u^2 \cos 	heta \sin 	heta}{g \cos^2 \alpha} - \frac{g \s

Vedclass · Accepted Answer

$20$

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(D) The range $R$ of a projectile on an inclined plane is given by the formula:$R = \frac{2 u^2 \cos 	heta \sin 	heta}{g \cos^2 \alpha} - \frac{g \sin \alpha}{2} \left( \frac{2 u \sin 	heta}{g \cos \alpha} ight)^2$Simplifying this,we get:$R = \frac{2 u^2 \cos 	heta \sin 	heta}{g \cos^2 \alpha} - \frac{2 u^2 \sin^2 	heta \sin \alpha}{g \cos^2 \alpha}$$R = \frac{2 u^2 \sin 	heta}{g \cos^2 \alpha} (\cos 	heta \cos \alpha - \sin 	heta \sin \alpha)$$R = \frac{2 u^2 \sin 	heta \cos(	heta + \alpha)}{g \cos^2 \alpha}$Given $u = 2 \, m/s$,$	heta = 15^o$,$\alpha = 30^o$,and $g = 10 \, m/s^2$:$R = \frac{2 	imes (2)^2 	imes \sin(15^o) 	imes \cos(45^o)}{10 	imes \cos^2(30^o)}$$R = \frac{8 	imes \sin(15^o) 	imes \frac{1}{\sqrt{2}}}{10 	imes (\frac{\sqrt{3}}{2})^2} = \frac{8 	imes \frac{\sqrt{3}-1}{2\sqrt{2}} 	imes \frac{1}{\sqrt{2}}}{10 	imes \frac{3}{4}}$$R = \frac{8 	imes \frac{\sqrt{3}-1}{4}}{7.5} = \frac{2(\sqrt{3}-1)}{7.5} = \frac{2(1.732-1)}{7.5} = \frac{1.464}{7.5} \approx 0.195 \, m = 19.5 \, cm \approx 20 \, cm$.

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