A) \[\sqrt{n\,g\,h}\]
B) \[\frac{1}{2}\sqrt{n\,g\,h}\]
C) \[\sqrt{2\,n\,g\,h}\]
D) \[\sqrt{\frac{1}{2}n\,g\,h}\]
Correct Answer: D
Solution :
Let \[u\] be the velocity of projection of the particle which is projected upwards from the bottom of the tower. Let the two particles meet after \[t\,\sec .\]Then when the particle falling downwards \[\frac{h}{n}=\frac{1}{2}g{{t}^{2}}\] ?(i) and when the particle projected upwards \[h-\frac{h}{n}=ut-\frac{1}{2}g{{t}^{2}}\] ?(ii) On adding Eqs. (i) and (ii), we get \[h=ut\Rightarrow t=\frac{h}{u}\] Putting the value of t in Eq. (i), we get \[\frac{h}{n}=\frac{1}{2}g\left( \frac{{{h}^{2}}}{{{u}^{2}}} \right)\]\[\Rightarrow \]\[{{u}^{2}}=\frac{1}{2}n\,g\,h\] \[\Rightarrow \] \[u=\sqrt{\frac{1}{2}\,n\,g\,h}\]
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