Question

Two towers AB and CD are situated at a distance $d$ apart. An object of mass $m$ is thrown from the top of AB horizontally with a velocity of $10m/s$ towards CD. Simultaneously another object of mass $2m$ is thrown from the top of CD at an angle ${60}^{\circ }$ with the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick with each other. Find the distance between the foot of the towers (approximately).

• A. $13m$
• B. $17m$
• C. $20m$
• D. $24m$

Answer 1

The correct option is B $17m$


Let $t$ be the time taken for collision. For mass $m$ thrown horizontally from $A$ for horizontal motion.
$PM=10t\dots \dots \left(1\right)$
For vertical motion $u_A=0,S_A=y,a_A=g$
$S_A=u_{y}t+\frac{1}{2}{a}_y{t}^2$ (Taking downward positive)
$\Rightarrow y=\frac{1}{2}g{t}^2\dots \dots \left(2\right)$
For mass $2\text{m}$ thrown from $C$, for horizontal motion
$QM=\left[10\cos {60}^{\circ }\right]t\Rightarrow QM=5t\dots \dots \left(3\right)$
For vertical motion:
Taking downward as positive
$u_B=10\sin {60}^{\circ }=5\sqrt{3}$
$a_y=g,S_B=y+10,S=u_{B}t+\frac{1}{2}a{t}^2$
$\Rightarrow y+10=5\sqrt{3}t+\frac{1}{2}g{t}^2\dots \dots \left(4\right)$
From equations (2) and (4)
$\frac{1}{2}g{t}^2+10=5\sqrt{3}t+\frac{1}{2}g{t}^2$
$t=\frac{2}{\sqrt{3}}s$
$BD=PM+MQ$
$=10t+10\cos {60}^{\circ }t$ (Total horizontal distance covered by both the particles)
$=10t+5t=15t=15\times \frac{2}{\sqrt{3}}$
$=10\sqrt{3}=17.32m$

Alternative solution:

$u_{AC}=[u_A-(-u_C\cos {60}^{\circ }\left)\right]\hat{i}+[0-(-u_c\sin {60}^{\circ }\left)\right]\hat{j}$
$\Rightarrow u_{AC}=15\hat{i}+5\sqrt{3}\hat{j}$
$\tan \theta =\frac{CE}{AE}$
$\Rightarrow \tan \theta =\frac{10}{d}$
Also $\tan \theta =\frac{u_{ACx}}{u_{ACy}}$
$\Rightarrow \tan \theta =\frac{0+u_C\sin {60}^{\circ }}{u_A+u_C\cos {60}^{\circ }}$
$\Rightarrow \frac{10}{d}=\frac{(10\times \frac{\sqrt{3}}{2})}{10+10\times \frac{1}{2}}$
$\Rightarrow d=10\sqrt{3}m$