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 |
| \[h=ut+\frac{1}{2}g{{t}^{2}}\] |
| or \[h=0+\frac{1}{2}g{{t}^{2}}\] |
| \[\therefore \] \[\frac{{{h}_{1}}}{{{h}_{2}}}={{\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}\] |
| Given, \[{{h}_{1}}=16\,m,\,{{h}_{2}}=25\,m\] |
| \[\therefore \] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{h}_{1}}}{{{h}_{2}}}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\] |
| Note: Time taken by the object in falling does not depend on mass of object. |
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