Question

Two particles $A$ and $B$ are projected simultaneously from the top of two towers of height $10\text{m}$ and $20\text{m}$ respectively. Particle $A$ is projected upwards at an angle of ${45}^{\circ }$ with horizontal at a speed of $10\sqrt{2}\text{m/s}$ and particle $B$ is projected horizontally at a speed of $10\text{m/s}$ towards $A$. What is the distance between towers if particles collide in the air ?Take $g=10{\text{m/s}}^2$.

• A. $10.5\text{m}$
• B. $25\text{m}$
• C. $20\text{m}$
• D. $35\text{m}$

Answer 1

The correct option is C $20\text{m}$

The velocity components of $A$,
$(v_A{)}_x=10\sqrt{2}\times \sin {45}^{\circ }=10\text{m/s}$
$(v_A{)}_x=10\sqrt{2}\times \sin {45}^{\circ }=10\text{m/s}$
Similarly,
The velocity components of $B$,
$(v_B{)}_x=-10\text{m/s}$
$(v_B{)}_y=0$
Relative acceleration $\Rightarrow a_{AB}=a_A-a_B=-g-(-g)=0$
When $A$ collides with $B$, the displacement of $A$ w.r.t $B$ is,
$S_y=+10\text{m}$
$\Rightarrow$The velocity of $A$ w.r.t $B$ in $y$ direction is
$(v_{AB}{)}_y=(v_A{)}_y-(v_B{)}_y=10-0$
$\therefore (v_{AB}{)}_y=10\text{m/s}$
$\Rightarrow$Time taken to collide will be
$t=\frac{S_y}{(v_{AB}{)}_y}$
$t=\frac{10}{10}=1\text{s}$
Similarly, The velocity of $A$ w.r.t $B$ in $y$ direction is
$(v_{AB}{)}_x=(v_A{)}_x-(v_B{)}_x=10-(-10)$
$\Rightarrow (v_{AB}{)}_x=20\text{m/s}$

Hence relative separation in horizontal (Distance between towers) is given by:
$d=(v_{AB}{)}_x\times t$
$\therefore d=20\times 1=20\text{m}$