A) 5/4
B) 12/5
C) 5/12
D) 4/5
Correct Answer: D
Solution :
As bodies are dropped from a certain height, their initial velocities are zero, i.e., \[u=0.\] Using \[h=ut+\frac{1}{2}g{{t}^{2}}\]or \[h=0+\frac{1}{2}g{{t}^{2}}\] \[\frac{{{h}_{1}}}{{{h}_{2}}}={{\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}\] Given, \[{{h}_{1}}=16\,m,\,{{h}_{2}}=25\,m\] \[\Rightarrow \]\[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{h}_{1}}}{{{h}_{2}}}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\]
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