Question

A particle is projected on an inclined plane with a speed u as shown in Fig. Find the range of the particle on the inclined plane.

• A. $=\frac{3u^2(\sqrt{3}-1)}{3g}$
• B. $=\frac{2u^2(\sqrt{3}-1)}{3g}$
• C. $=\frac{4u^2(\sqrt{3}-1)}{3g}$
• D. $=\frac{5u^2(\sqrt{3}-1)}{3g}$

Answer 1

Answer: (B)

$=\frac{2u^2(\sqrt{3}-1)}{3g}$

$R=\frac{u^2}{g{\cos }^2{\theta }_0}\left[\sin \right(2\alpha +{\theta }_0)-\sin {\theta }_0]$

where ${\theta }_0={30}^o$ and $\alpha ={45}^o$

Then, $R=\frac{u^2}{g{\cos }^2{30}^o}\left[\sin \right(2\times {45}^o+{30}^o)-\sin {30}^o]$

$=\frac{2u^2(\sqrt{3}-1)}{3g}$