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Question

(A) Let the initial velocity be $\vec{u} = u \cos 	heta \hat{i} + u \sin 	heta \hat{j}$.After time $t$,the velocity vector is $\vec{v} = u \cos 	he

Vedclass · Accepted Answer

$u / (g \sin 	heta)$

Vedclass · Answer

(A) Let the initial velocity be $\vec{u} = u \cos 	heta \hat{i} + u \sin 	heta \hat{j}$.After time $t$,the velocity vector is $\vec{v} = u \cos 	heta \hat{i} + (u \sin 	heta - gt) \hat{j}$.Given that the velocity vector $\vec{v}$ is perpendicular to the initial velocity $\vec{u}$,their dot product must be zero: $\vec{v} \cdot \vec{u} = 0$.$(u \cos 	heta \hat{i} + (u \sin 	heta - gt) \hat{j}) \cdot (u \cos 	heta \hat{i} + u \sin 	heta \hat{j}) = 0$.$u^2 \cos^2 	heta + (u \sin 	heta - gt)(u \sin 	heta) = 0$.$u^2 \cos^2 	heta + u^2 \sin^2 	heta - ugt \sin 	heta = 0$.$u^2 (\cos^2 	heta + \sin^2 	heta) = ugt \sin 	heta$.$u^2 = ugt \sin 	heta$.$t = \frac{u}{g \sin 	heta}$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

$A$ particle is projected from the ground at an angle of $\theta$ with the horizontal with an initial speed of $u$. The time after which the velocity vector of the projectile is perpendicular to the initial velocity is:

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