Projectile Motion – Perpendicular Velocity – Rankers Physics

Ground to Ground Projectile: Practice Problem & Solution

Assertion (A): A particle is projected from ground on a horizontal plane with speed \(10\text{ ms}^{-1}\) and angle of projection \(37^\circ\) with horizontal. Its velocity vector will be perpendicular to initial velocity vector after \(\frac{4}{3}\text{ s}\). Reason (R): Two vectors \(\vec{v}\) and \(\vec{u}\) are perpendicular then \(\vec{u} \cdot \vec{v} = 0\).
(1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are false

Solution Explained:

To solve this problem, we apply the core principles of Ground to Ground Projectile. Understanding the underlying formula is key to arriving at the correct answer below:

Assertion (A): Initial velocity \(\vec{u} = (10 cos 37^\circ)\hat{i} + (10 sin 37^\circ)\hat{j} approx 8\hat{i} + 6\hat{j}\). Velocity at time (t) is \(\vec{v}(t) = 8\hat{i} + (6 - gt)\hat{j}\). For perpendicularity, \(\vec{u} \cdot \vec{v} = 0 ⇒ 64 + 6(6 - gt) = 0 ⇒ 100 - 6gt = 0\). With \(g=10\text{ m/s}^2), (t = 100/60 = 5/3\text{ s}\). The assertion states \(4/3\text{ s})\, so (A) is False.
Reason (R): The dot product of two perpendicular vectors is indeed zero. So (R) is True.
Since (A) is false and (R) is true, none of the given options are strictly correct. However, if (A) is false, options (1), (2), (3) are ruled out, leaving (4) by elimination, despite (R) being true.

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