A) 5/4
B) 12/5
C) 5/12
D) 4/5
Correct Answer: D
Solution :
Key Idea: As bodies are dropped from a certain height, their initial velocities are zero ie,\[u=0\]. For free fall from a height\[u=0\](initial velocity). From second equation of motion \[h=ut+\frac{1}{2}g{{t}^{2}}\] Or \[h=0+\frac{1}{2}g{{t}^{2}}\] \[\therefore \] \[\frac{{{h}_{1}}}{{{h}_{2}}}={{\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}\] Given, \[{{h}_{1}}=16m,{{h}_{2}}=25m\] \[\therefore \] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{h}_{1}}}{{{h}_{2}}}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\] NOTE: Time taken by the object in falling does not depend on mass of object.
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