Physics - Kinematics Question with Solution | TestHub

Since both rocks have the same launch speed and the same range $d$, their launch angles must give the same value of $\text{sin}2\theta$; as the angles are different and one is $37^o$, the other is $53^o$. The rocks are launched simultaneously, so their relative acceleration is zero and their relative velocity ${\overrightarrow{v}}_{BA}={\overrightarrow{v}}_B-{\overrightarrow{v}}_A$ is constant. Taking the components (with $\text{cos}37^o=3/5,\text{sin}37^o=4/5$),

$${\overrightarrow{v}}_{BA}=v\left(-\text{cos}53^o-\text{cos}37^o\right)\hat{i}+v\left(\text{sin}53^o-\text{sin}37^o\right)\hat{j}=v\left(-\frac{7}{5}\hat{i}-\frac{1}{5}\hat{j}\right)$$

so $\left|{\overrightarrow{v}}_{BA}\right|=\frac{v}{5}\sqrt{49+1}=v\sqrt{2}$, directed at an angle of about $53^o-45^o\approx {8.1}^o$ below the horizontal line $BA$.

The minimum separation equals the component of the initial separation $d$ perpendicular to ${\overrightarrow{v}}_{BA}$:

$$x_{\text{min}}=d\text{sin}(53^o-45^o)\approx \frac{d}{5\sqrt{2}}=\frac{10\sqrt{2}}{5\sqrt{2}}=2\text{m}$$