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 i.e., u = 0. For free fall from a height u = 0 (initial velocity). From second equation of motion \[h=ut+\frac{1}{2}g{{t}^{2}}\] \[orh=0+\frac{1}{2}g{{t}^{2}}\] \[\therefore \frac{{{h}_{1}}}{{{h}_{2}}}={{\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}\] \[Given,{{h}_{1}}=16\,m,\,{{h}_{2}}=25\,m\] \[\therefore \frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{h}_{1}}}{{{h}_{2}}}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\] Note: Time taken by the object in falling does not depend on mass of object.
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