A) \[\frac{25}{16}\]
B) \[\frac{5}{4}\]
C) \[\frac{4}{5}\]
D) \[\frac{16}{25}\]
Correct Answer: C
Solution :
\[h=0+\frac{1}{2}g{{t}^{2}}\Rightarrow {{t}^{2}}\propto h\] \[\therefore \] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{h}_{1}}}{{{h}_{2}}}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\]
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