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. 30⁰
• B. 45⁰
• C. 60⁰
• D. 90⁰

Answer 1

Answer: (A)

30⁰

Answer 2

Maximum range up the inclined plane

$({Rup\frac{u^{2}}{g(1 + \sin\alpha)}}_{\max}$

Maximum range down the inclined plane

$({Rdown\frac{u^{2}}{g(1 - \sin\alpha)}}_{\max}$

and according to problem : $\frac{u^{2}}{g(1 - \sin\alpha)} = 3 \times \frac{u^{2}}{g(1 + \sin\alpha)}$

By solving α = 30⁰