Question

Two particles $\text{A}$ and $\text{B}$ are projected simultaneously from two towers of height $10\text{m}$ and $20\text{m}$ respectively. Particle $\text{A}$ is projected with an initial speed of $10\sqrt{2}\text{m/s}$ at an angle of ${45}^{\circ }$ with horizontal, while particle $\text{B}$ is projected horizontally with speed $10\text{m/s}$. If they collide in air, what is the distance $d$ between the towers and time taken for collision?
$(g=10{\text{m/s}}^2)$

• A. $18\text{m},3\text{s}$
• B. $20\text{m},1\text{s}$
• C. $16\text{m},8\text{s}$
• D. $15\text{m},2\text{s}$

Answer 1

The correct option is B $20\text{m},1\text{s}$


Initial relative velocity of particle $\text{B}$ with respect to $\text{A}$,

${\overrightarrow{v}}_{\text{BA}}={\overrightarrow{v}}_{\text{B}}-{\overrightarrow{v}}_{\text{A}}$
${\overrightarrow{v}}_{\text{BA}}=[-10\hat{i}-(10\hat{i}+10\hat{j}\left)\right]\text{m/s}$
${\overrightarrow{v}}_{\text{BA}}=(-20\hat{i}-10\hat{j})\text{m/s}$

Relative acceleration of particle $\text{B}$ with respect to $\text{A}$,

${\overrightarrow{a}}_{\text{BA}}={\overrightarrow{a}}_{\text{B}}-{\overrightarrow{a}}_{\text{A}}=-g\hat{j}-(-g\hat{j})=-10\hat{j}-(-10\hat{j})=0$

Applying the equation of relative motion,

$s_{\text{rel}}=v_{\text{rel}}\times t[$considering particle $\text{A}$ to be at rest$]$

Along $y-$axis,

$s_y=(v_{\text{AB}}{)}_y\times t$

From diagram, $s_y=-10\text{m}$ and from above equation, $(v_{\text{AB}}{)}_y=-10\text{m/s}$

$\Rightarrow -10=-10t$

$\Rightarrow t=1\text{s}.....\left(1\right)$

Similarly, along $x-$axis,

$-d=-20t$

$\Rightarrow d=20t$

$\therefore d=20\text{m}[$from $\left(1\right),t=1\text{s}]$

Hence, option $\left(\text{A}\right)$ is the correct answer.