Question

A projectile is fired at an angle $\theta$ with the horizontal. Find the condition under which it lands perpendicular on an inclined plane inclination $\alpha$ as shown in figure.

• A. $\sin\alpha = \cos (\theta - \alpha)$
• B. $\cos\alpha = \sin(\theta - \alpha)$
• C. $\tan\alpha = \cot(\theta - \alpha)$
• D. $\cot(\theta - \alpha) = 2\tan\alpha$

Answer 1

Answer: (D)

$\cot(\theta - \alpha) = 2\tan\alpha$

Answer 2

For a projectile landing perpendicularly on an inclined plane (inclination $\alpha$), the condition derived by equating the time based on the motion perpendicular to the plane and along the incline is:

$$\cot(\theta-\alpha) = 2\tan\alpha.$$

Explanation:

1. Resolve the projectile motion along and perpendicular to the incline.
2. Write the equation of motion perpendicular to the incline and determine the time of flight $T$.
3. Write the velocity condition (zero final velocity component along the inclined direction) using the corresponding acceleration component.
4. Equate the two expressions for $T$ to get $\cot(\theta-\alpha)=2\tan\alpha$.