The correct answer is: The direction of projection should bisect the angle between the inclined plane and the vertical for a range of a projectile on inclined plane to be maximum.
The range of a projectile is the maximum horizontal distance that it travels. The range of a projectile on an inclined plane is given by the following equation:
$$R = \frac{v^2 \sin 2\theta}{g \cos \theta}$$
where $v$ is the initial velocity of the projectile, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.
The maximum range occurs when $\theta = 45^\circ$. This is because when $\theta = 45^\circ$, the horizontal component of the velocity is equal to the vertical component of the velocity. This means that the projectile will travel a maximum distance horizontally before it hits the ground.
When the direction of projection bisects the angle between the inclined plane and the vertical, the angle of projection is equal to $45^\circ$. Therefore, the range of the projectile will be maximum.
The other options are incorrect because:
- Option A is incorrect because the range of the projectile is not zero when the direction of projection bisects the angle between the inclined plane and the vertical.
- Option B is incorrect because the range of the projectile is not minimum when the direction of projection bisects the angle between the inclined plane and the vertical.
- Option C is incorrect because the range of the projectile is not maximum when the direction of projection is not bisecting the angle between the inclined plane and the vertical.