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Question

(A) Let the horizontal direction be the $x$-axis and the vertical direction be the $y$-axis.At time $t$,the velocity vector of the first particle is $

Vedclass · Accepted Answer

$\frac{\sqrt{u_1 u_2}}{g}$

Vedclass · Answer

(A) Let the horizontal direction be the $x$-axis and the vertical direction be the $y$-axis.At time $t$,the velocity vector of the first particle is $\vec{v}_1 = u_1 \hat{i} - gt \hat{j}$.The velocity vector of the second particle is $\vec{v}_2 = -u_2 \hat{i} - gt \hat{j}$.Since the velocities are mutually perpendicular,their dot product must be zero: $\vec{v}_1 \cdot \vec{v}_2 = 0$.$(u_1 \hat{i} - gt \hat{j}) \cdot (-u_2 \hat{i} - gt \hat{j}) = 0$.$-u_1 u_2 + g^2 t^2 = 0$.$g^2 t^2 = u_1 u_2$.$t = \frac{\sqrt{u_1 u_2}}{g}$.

Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$

Two particles are simultaneously projected in opposite directions horizontally from a given point in space,where gravity $g$ is uniform. If $u_1$ and $u_2$ are their initial speeds,then the time $t$ after which their velocities are mutually perpendicular is given by

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Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed $(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed $(b)$ $0^o$

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