A) \[1:\sqrt{2}\]
B) \[\sqrt{2}:1\]
C) \[1:2\]
D) \[2:1\]
Correct Answer: A
Solution :
From relation\[h=ut+\frac{1}{2}g{{t}^{2}}\] (with\[u=0,\]) we have \[h=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \] \[t=\sqrt{\left( \frac{2h}{g} \right)}\propto \sqrt{h}\] \[\therefore \] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\left( \frac{{{h}_{1}}}{{{h}_{2}}} \right)}\sqrt{\left( \frac{h}{2h} \right)}=\frac{1}{\sqrt{2}}\]
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