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(A) Let the first body be $A$ and the second body be $B$. Both are thrown from the origin $(0,0)$ at $t=0$.For body $A$ (thrown vertically up): $\vec{

Vedclass · Accepted Answer

$20 \ m$

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(A) Let the first body be $A$ and the second body be $B$. Both are thrown from the origin $(0,0)$ at $t=0$.For body $A$ (thrown vertically up): $\vec{v}_A = v_0 \hat{j}$, $\vec{a}_A = -g \hat{j}$.Position of $A$ at time $t$: $\vec{r}_A = (v_0 t) \hat{j} - \frac{1}{2} g t^2 \hat{j}$.For body $B$ (thrown at $60^{\circ}$ to the vertical, which is $30^{\circ}$ to the horizontal): $\vec{v}_B = v_0 \sin 60^{\circ} \hat{i} + v_0 \cos 60^{\circ} \hat{j} = v_0 \cos 30^{\circ} \hat{i} + v_0 \sin 30^{\circ} \hat{j}$.Position of $B$ at time $t$: $\vec{r}_B = (v_0 \cos 30^{\circ} t) \hat{i} + (v_0 \sin 30^{\circ} t - \frac{1}{2} g t^2) \hat{j}$.The relative position vector $\vec{r}_{AB} = \vec{r}_B - \vec{r}_A = (v_0 \cos 30^{\circ} t) \hat{i} + (v_0 \sin 30^{\circ} t - v_0 t) \hat{j}$.Substituting the values $v_0 = 10 \ m \ s^{-1}$ and $t = 2 \ s$:$\vec{r}_{AB} = (10 	imes \frac{\sqrt{3}}{2} 	imes 2) \hat{i} + (10 	imes \frac{1}{2} 	imes 2 - 10 	imes 2) \hat{j} = (10\sqrt{3}) \hat{i} + (10 - 20) \hat{j} = 10\sqrt{3} \hat{i} - 10 \hat{j}$.The distance is the magnitude of $\vec{r}_{AB}$:$|\vec{r}_{AB}| = \sqrt{(10\sqrt{3})^2 + (-10)^2} = \sqrt{300 + 100} = \sqrt{400} = 20 \ m$.

Two bodies were thrown simultaneously from the origin: one straight up and the other at an angle of $60^{\circ}$ to the vertical. The initial velocity of each body is $10 \ m \ s^{-1}$. Neglecting air resistance, the distance between the two bodies after $t=2 \ s$ is (Use $g=10 \ m \ s^{-2}$):

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