Question

From the top of two towers of height $29\text{m}$ and $20\text{m}$, two balls $\text{A}$ and $\text{B}$ are projected simultaneously with velocities $2\sqrt{2}\text{m/s}$ and $14\sqrt{2}\text{m/s}$ towards each other at an angle of inclination of ${45}^{\circ }$ each as shown in figure. Find the minimum separation between the two balls

• A. zero
• B. $2\text{m}$
• C. $4\text{m}$
• D. $6\text{m}$

Answer 1

The correct option is D $6\text{m}$

The horizontal velocity of $\text{B}$ wrt $\text{A}$ is given by,

$(V_{BA}{)}_x=14\sqrt{2}\cos {45}^{\circ }+2\sqrt{2}\cos {45}^{\circ }=16$

The vertical velocity of $\text{B}$ wrt $\text{A}$ is given by,

$(V_{BA}{)}_y=14\sqrt{2}\sin {45}^{\circ }-2\sqrt{2}\sin {45}^{\circ }=12$





In triangle $\text{ABC}$, we have:

$\tan {37}^{\circ }=\frac{AC}{BC}=\frac{AC}{22}$

$\Rightarrow AC=16.5\text{m}$

Clearly, the minimum seperation between the two ball is given by,

$d=\text{AQ}\sin {53}^{\circ }$

$\text{AQ}=16.5-9=7.5\text{m}$

$\Rightarrow d=7.5\sin {53}^{\circ }$

$\Rightarrow d=6\text{m}$

Hence, option $\left(D\right)$ is the correct answer.