Two particles are projected from a tower in o | vedclass

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Question

(A) Let the two particles be $1$ and $2$. Both are projected horizontally with $u_1 = 10\,m/s$ (positive x-direction) and $u_2 = -10\,m/s$ (negative x

Vedclass · Accepted Answer

$(A \rightarrow p, B \rightarrow s, C \rightarrow s, D \rightarrow p)$

Vedclass · Answer

(A) Let the two particles be $1$ and $2$. Both are projected horizontally with $u_1 = 10\,m/s$ (positive x-direction) and $u_2 = -10\,m/s$ (negative x-direction).$(A)$ Relative acceleration: Both particles are under gravity,so $\vec{a}_1 = \vec{a}_2 = -g\hat{j}$. Thus,$\vec{a}_{rel} = \vec{a}_1 - \vec{a}_2 = 0$. Matches $(p)$.$(B)$ Relative velocity: $\vec{v}_1 = (10\hat{i} - gt\hat{j})$ and $\vec{v}_2 = (-10\hat{i} - gt\hat{j})$. $\vec{v}_{rel} = \vec{v}_1 - \vec{v}_2 = 20\hat{i}$. Magnitude is $20\,m/s$. Matches $(s)$.$(C)$ Horizontal distance: $x_1 = 10t$ and $x_2 = -10t$. Distance $d_x = |x_1 - x_2| = |10t - (-10t)| = 20t$. At $t=1\,s$,$d_x = 20\,m$. Matches $(s)$.$(D)$ Vertical distance: $y_1 = -\frac{1}{2}gt^2$ and $y_2 = -\frac{1}{2}gt^2$. Distance $d_y = |y_1 - y_2| = 0$. Matches $(p)$.Correct matching: $(A \rightarrow p, B \rightarrow s, C \rightarrow s, D \rightarrow p)$.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

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