Question

A projectile is thrown into air with velocity u at an angle $\theta$ to the horizontal. The time at which its direction of motion is perpendicular to its initial direction is

• A. $\frac{u}{g \sin \theta}$
• B. $\frac{u}{g \cos \theta}$
• C. $\frac{u}{g \tan \theta}$
• D. $\frac{u}{g \cot \theta}$

Hint:

This condition only occurs if the initial angle $\theta$ is such that the time $t$ is less than the total time of flight ($T = \frac{2u \sin \theta}{g}$).

Answer 1

The Correct Option is A

Initial velocity vector: $\vec{u} = u \cos \theta \hat{i} + u \sin \theta \hat{j}$. Velocity at time $t$: $\vec{v} = u \cos \theta \hat{i} + (u \sin \theta - gt) \hat{j}$. For the motion to be perpendicular, the dot product $\vec{u} \cdot \vec{v} = 0$: $$(u \cos \theta)(u \cos \theta) + (u \sin \theta)(u \sin \theta - gt) = 0$$ $$u^2 \cos^2 \theta + u^2 \sin^2 \theta - ugt \sin \theta = 0$$ Since $\cos^2 \theta + \sin^2 \theta = 1$: $$u^2 - ugt \sin \theta = 0$$ $$u = gt \sin \theta$$ $$t = \frac{u}{g \sin \theta}$$ This time $t$ is when the velocity vector makes an angle of $(90+\theta)$ with the horizontal.