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 find the angle of inclination of the inclined plane.
A. $\beta ={{40}^{0}}$
B. $\beta ={{20}^{0}}$
C. $\beta ={{60}^{0}}$
D. $\beta ={{30}^{0}}$

Answer 1

The maximum and minimum range ( ${{R}_{u}}\text{ and }{{R}_{d}}$ ) for projectile motion up and down the incline respectively can be written in terms of the initial velocity and angle of inclination of the plane. Using the given condition (${{R}_{d}}=3{{R}_{u}}$ ), the angle of inclination can be calculated.

Formula used:
Maximum range for projectile motion up the incline
${{R}_{u}}=\dfrac{{{u}^{2}}}{g(1+sin\beta )}$
Maximum range for projectile motion down the incline
${{R}_{d}}=\dfrac{{{u}^{2}}}{g(1-sin\beta )}$
Here
$u$ is the velocity of projection
$g$ is the acceleration due to gravity
$\beta$ is the angle of inclination of the inclined plane

Complete step by step solution:
The maximum range up the plane is three times the maximum range down the plane.

& {{R}_{d}}=3{{R}_{u}} \\\ & \dfrac{{{u}^{2}}}{g(1-sin\beta )}=\dfrac{3{{u}^{2}}}{g(1+sin\beta )} \\\ & \\\ & (1+\sin \beta )=3(1-\sin \beta ) \\\ & \sin \beta =\dfrac{1}{2} \\\ & \beta ={{30}^{0}} \\\ \end{aligned}$$ **The correct answer is option D.** **Note:** The angle of projection for maximum range up the incline is $\dfrac{\pi }{4}-\dfrac{\beta }{2}$ and the angle of projection for maximum range down the incline is $\dfrac{\pi }{4}+\dfrac{\beta }{2}$.