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(B) Let the initial velocity be $\vec{u} = u \cos \alpha \hat{i} + u \sin \alpha \hat{j}$.Let the velocity at point $P$ be $\vec{v} = v_x \hat{i} + v_

Vedclass · Accepted Answer

$u \cot \alpha$

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(B) Let the initial velocity be $\vec{u} = u \cos \alpha \hat{i} + u \sin \alpha \hat{j}$.Let the velocity at point $P$ be $\vec{v} = v_x \hat{i} + v_y \hat{j}$.Since horizontal velocity remains constant,$v_x = u \cos \alpha$.Given that the velocity $\vec{v}$ is perpendicular to the initial velocity $\vec{u}$,their dot product must be zero:$\vec{u} \cdot \vec{v} = 0$$(u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) \cdot (u \cos \alpha \hat{i} + v_y \hat{j}) = 0$$u^2 \cos^2 \alpha + u \sin \alpha v_y = 0$$v_y = -\frac{u^2 \cos^2 \alpha}{u \sin \alpha} = -u \frac{\cos^2 \alpha}{\sin \alpha}$The magnitude of the velocity $v$ is given by $v = \sqrt{v_x^2 + v_y^2}$.$v = \sqrt{(u \cos \alpha)^2 + (-u \frac{\cos^2 \alpha}{\sin \alpha})^2}$$v = u \cos \alpha \sqrt{1 + \frac{\cos^2 \alpha}{\sin^2 \alpha}} = u \cos \alpha \sqrt{\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha}}$$v = u \cos \alpha \cdot \frac{1}{\sin \alpha} = u \cot \alpha$.

Column $I$	Column $II$
$(A)$ Displacement	$(p)$ $8 \sin 2$
$(B)$ Distance	$(q)$ $4$
$(C)$ Average velocity	$(r)$ $2 \sin 2$
$(D)$ Average acceleration	$(s)$ $4 \sin 2$

$A$ body is projected with velocity $u$ making an angle $\alpha$ with the horizontal. Its velocity when it is perpendicular to the initial velocity vector $u$ is

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