Question

For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

Maximum range up the inclined plane $(R_{max}{)}_{up}=\frac{u^2}{g(1+sin\alpha )}$
Maximum range down the inclined plane $(R_{max}{)}_{down}=\frac{u^2}{g(1-sin\alpha )}$
and according to problem : $\frac{u^2}{g(1-sin\alpha )}=3\times \frac{u^2}{g(1+sin\alpha )}$
By solving $\alpha ={30}^{\circ }$