There are two points P and Q on a projectile | vedclass

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Question

(D) Let the initial velocity of the projectile be $u$ at an angle $	heta$ with the horizontal. The horizontal component of velocity remains constant

Vedclass · Accepted Answer

$\cot \alpha = \frac{v_Q}{v_P}$

Vedclass · Answer

(D) Let the initial velocity of the projectile be $u$ at an angle $	heta$ with the horizontal. The horizontal component of velocity remains constant throughout the motion,$v_x = u \cos 	heta$.At any point,the velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j}$.For point $P$,$\vec{v}_P = v_x \hat{i} + v_{yP} \hat{j}$. The angle $\alpha$ with the horizontal is given by $	an \alpha = \frac{v_{yP}}{v_x}$,so $v_{yP} = v_x 	an \alpha$.For point $Q$,$\vec{v}_Q = v_x \hat{i} + v_{yQ} \hat{j}$.Since $\vec{v}_P \perp \vec{v}_Q$,their dot product is zero: $\vec{v}_P \cdot \vec{v}_Q = 0$.$(v_x \hat{i} + v_{yP} \hat{j}) \cdot (v_x \hat{i} + v_{yQ} \hat{j}) = 0 \implies v_x^2 + v_{yP} v_{yQ} = 0$.Thus,$v_{yQ} = -\frac{v_x^2}{v_{yP}} = -\frac{v_x^2}{v_x 	an \alpha} = -v_x \cot \alpha$.The magnitudes are $v_P = \sqrt{v_x^2 + v_{yP}^2} = \sqrt{v_x^2 + v_x^2 	an^2 \alpha} = v_x \sec \alpha$.And $v_Q = \sqrt{v_x^2 + v_{yQ}^2} = \sqrt{v_x^2 + v_x^2 \cot^2 \alpha} = v_x \csc \alpha$.Therefore,$\frac{v_Q}{v_P} = \frac{v_x \csc \alpha}{v_x \sec \alpha} = \frac{1/\sin \alpha}{1/\cos \alpha} = \cot \alpha$.

There are two points $P$ and $Q$ on a projectile path with velocities $v_P$ and $v_Q$ respectively,such that $v_P$ is perpendicular to $v_Q$. If $\alpha$ is the angle that $v_P$ makes with the horizontal at point $P$,find the correct option.

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