Projectile Range on Inclined Plane – Rankers Physics
Topic: Kinematics
Subtopic: Ground to Ground Projectile

Projectile Range on Inclined Plane


Assertion (A): The maximum range along the inclined plane, when thrown downward is greater than that when thrown upward along the same inclined plane with same speed at same angle from incline.
Reason (R): The maximum range along inclined plane is independent of angle of inclination.
 
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution:

The maximum range down an incline is \(R_{\text{max, down}} = \frac{u^2}{g(1-\sin\alpha)}\) and up an incline is \(R_{\text{max, up}} = \frac{u^2}{g(1+\sin\alpha)}\). Since \(1-\sin\alpha 0\), \(R_{\text{max, down}} > R_{\text{max, up}}\). So (A) is true. Both formulas clearly depend on the angle of inclination \(\alpha\). Thus (R) is false.

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