Projectile Motion – Perpendicular Velocity – Rankers Physics
Topic: Kinematics
Subtopic: Ground to Ground Projectile

Projectile Motion – Perpendicular Velocity

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:

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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