Question

Two particles A and B are projected simultaneously from the top of two towers as shown in the figure. At what height from the ground will they collide? (Take $g=10{\text{ms}}^{-2}$)

• A. 76 m
• B. 174 m
• C. 92 m
• D. 158 m

Answer 1

The correct option is D 158 m

Taking downward as positive direction. For the particles to collide Relative displacement in y- direction should be '0'. So,
Thus $(S_{rel}{)}_y=0$ and $(S_{rel}{)}_x=150$
$(S_{rel}{)}_x=(u_{rel}{)}_{x}t+\frac{1}{2}(a_{rel}{)}_x{t}^2$
$150=(45\sin {37}^{\circ }+60\sin {53}^{\circ })t$ $(\because (a_{rel}{)}_x=0)$
$150=(24+48)t$
$\Rightarrow t=2\text{sec}$
Displacement in y-direction for $A$,
$S=u_{Ay}t+\frac{1}{2}{a}_{Ay}{t}^2$
$S=45\cos {37}^{\circ }t+\frac{1}{2}g{t}^2$
$\Rightarrow S=36t+\frac{1}{2}\times 10\times t^2$
At $t=2s$, distance covered by $A$ in vertical direction
$=S=72+20\text{m}=92\text{m}$
$\therefore$ Height from the ground $=250-92=158\text{m}$