| Two particles are thrown from top of a tower from same point in opposite directions as shown below. |
 |
| Their velocities are perpendicular after a time interval of |
A) \[\sqrt{3}s\]
B) \[\frac{\sqrt{3}}{2}s\]
C) \[\frac{\sqrt{3}}{5}s\]
D) \[\frac{\sqrt{3}}{7}s\]
Correct Answer: C
Solution :
| Velocities of the particle after time (seconds is |
| \[{{{v}'}_{1}}={{v}_{1}}+g{{t}_{1}}\] |
| \[{{{v}'}_{2}}={{v}_{2}}+g{{t}_{2}}\] |
| \[{{{v}'}_{1}}\bot {{{v}'}_{2}}={{{v}'}_{1}}.{{{v}'}_{2}}=0\] |
| As, \[{{\upsilon }_{1}}\] and \[{{\upsilon }_{2}}\] are in opposite directions |
| \[\therefore \,\,\,\,{{v}_{1}}.{{v}_{2}}=-{{v}_{1}}{{v}_{2}}\] |
| \[\left( {{{{v}'}}_{1}}+g{{{{t}'}}_{1}} \right).\left( {{{{v}'}}_{2}}+g{{t}_{2}} \right)=0\]\[\Rightarrow -{{\upsilon }_{1}}{{\upsilon }_{2}}+{{g}^{2}}{{t}^{2}}=0\]\[\Rightarrow \,\,\,\,t=\frac{\sqrt{{{\upsilon }_{1}}{{\upsilon }_{2}}}}{g}=\frac{\sqrt{4\times 3}}{10}=\frac{\sqrt{3}}{5}\] |
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