question_answer

Two towers AB and CD are situated d distant apart.

AB is 20 m high and CD is 30 m high. From top of AB a particle of mass m is thrown towards CD with speed\[u=10m/s.\] At same instant, another particle of mass 2 m is thrown from CD towards AB at an angle of \[60{}^\circ \] with the horizontal in same vertical plane with same speed of \[10\,m{{s}^{-1}}.\]Distance between towers is

A) \[5\sqrt{3}m\]

B) \[6\sqrt{3}\,m\]

C) \[8\sqrt{3}\,m\]

D) \[10\sqrt{3}\,m\]

Correct Answer: D

Solution :

Let particles collide in mid air at sometime instant t after \[t=0.\]

Then, \[20-y=\frac{1}{2}g{{t}^{2}}\]

and \[30-y=10\sin 60{}^\circ t+\frac{1}{2}g{{t}^{2}}\]

[taking, B as origin]

Solving, we get \[t+\frac{1}{2}g{{t}^{2}}\]

Now, \[{{x}_{1}}=\]distance covered horizontally

by first particle \[=10t=\frac{20}{\sqrt{3}}m\]

\[{{x}_{2}}=\]horizontal distance of second particle

\[=10\cos 60{}^\circ t=\frac{10}{\sqrt{3}}m\]

\[\therefore \]Distance between towers,

\[d={{x}_{1}}+{{x}_{2}}=10\sqrt{3}m\]