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