Q134 / 180 · COMEDK 2026 (9 May Shift 1) · Physics
From the same point, two stones A and B are thrown simultaneously, A is thrown up vertically with a velocity of $10\ ms^{-1}$ and B is thrown up with the same velocity at $30^\circ$ with the horizontal. The separation between the balls at 2 s is
- Correct answerA$20\ \text{m}$
- B$10.5\ \text{m}$
- C$25\ \text{m}$
- D$22\ \text{m}$
Solution
Let the point of projection be the origin $(0,0)$.
Position vector of stone A at time $t$ is:
$\vec{r}_A = (10t - \dfrac{1}{2}gt^2)\hat{j}$
Position vector of stone B at time $t$ is:
$\vec{r}_B = (10\cos 30^\circ t)\hat{i} + (10\sin 30^\circ t - \dfrac{1}{2}gt^2)\hat{j}$
$\vec{r}_B = 5\sqrt{3}t\hat{i} + (5t - \dfrac{1}{2}gt^2)\hat{j}$
The relative position vector of A with respect to B is:
$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = -5\sqrt{3}t\hat{i} + 5t\hat{j}$
The separation between the stones is the magnitude of $\vec{r}_{AB}$:
$|\vec{r}_{AB}| = \sqrt{(-5\sqrt{3}t)^2 + (5t)^2}$
$|\vec{r}_{AB}| = \sqrt{75t^2 + 25t^2} = \sqrt{100t^2} = 10t$
At $t = 2\text{ s}$, the separation is:
$d = 10 \times 2 = 20\text{ m}$
Answer: $20\ \text{m}$
Position vector of stone A at time $t$ is:
$\vec{r}_A = (10t - \dfrac{1}{2}gt^2)\hat{j}$
Position vector of stone B at time $t$ is:
$\vec{r}_B = (10\cos 30^\circ t)\hat{i} + (10\sin 30^\circ t - \dfrac{1}{2}gt^2)\hat{j}$
$\vec{r}_B = 5\sqrt{3}t\hat{i} + (5t - \dfrac{1}{2}gt^2)\hat{j}$
The relative position vector of A with respect to B is:
$\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = -5\sqrt{3}t\hat{i} + 5t\hat{j}$
The separation between the stones is the magnitude of $\vec{r}_{AB}$:
$|\vec{r}_{AB}| = \sqrt{(-5\sqrt{3}t)^2 + (5t)^2}$
$|\vec{r}_{AB}| = \sqrt{75t^2 + 25t^2} = \sqrt{100t^2} = 10t$
At $t = 2\text{ s}$, the separation is:
$d = 10 \times 2 = 20\text{ m}$
Answer: $20\ \text{m}$
Hence, the correct answer is option: 'A'.