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(B) Let the initial velocity be $\vec{u} = u \cos \alpha \hat{i} + u \sin \alpha \hat{j}$.At point $P$,the velocity $\vec{v}$ is perpendicular to $\ve

Vedclass · Accepted Answer

$\frac{u \csc \alpha}{g}$

Vedclass · Answer

(B) Let the initial velocity be $\vec{u} = u \cos \alpha \hat{i} + u \sin \alpha \hat{j}$.At point $P$,the velocity $\vec{v}$ is perpendicular to $\vec{u}$,so $\vec{v} \cdot \vec{u} = 0$.The velocity at any time $t$ is $\vec{v} = u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}$.Taking the dot product: $(u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}) \cdot (u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) = 0$.$u^2 \cos^2 \alpha + u \sin \alpha (u \sin \alpha - gt) = 0$.$u^2 \cos^2 \alpha + u^2 \sin^2 \alpha - u \sin \alpha gt = 0$.$u^2 (\cos^2 \alpha + \sin^2 \alpha) = u \sin \alpha gt$.$u^2 = u \sin \alpha gt$.$t = \frac{u}{g \sin \alpha} = \frac{u \csc \alpha}{g}$.

$A$ particle is projected from point $O$ with velocity $u$ at an angle $\alpha$ with the horizontal. If at point $P$,its velocity is perpendicular to the initial direction of projection,then find the time taken to reach from $O$ to $P$.

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