A) \[12\,h\]
B) \[8\,h\]
C) \[9\,h\]
D) \[4\,h\]
Correct Answer: D
Solution :
From equation of motion, we have \[{{v}^{2}}={{u}^{2}}+2\,gh\] where u is initial velocity, v is final velocity and h is height of fall. Given, \[u=0\] (body starts from rest). \[\therefore \] \[{{v}^{2}}=2\,gh\] Given, \[{{v}_{1}}=v,\,\,{{v}_{2}}=2v,\,{{h}_{1}}=h\] \[\therefore \] \[\frac{{{v}_{1}}^{2}}{{{v}_{2}}^{2}}=\frac{2\,g{{h}_{1}}}{2\,g{{h}_{2}}}\] \[\Rightarrow \] \[\frac{{{v}^{2}}}{4{{v}^{2}}}=\frac{h}{{{h}_{2}}}\] \[\Rightarrow \] \[{{h}_{2}}=4\,h\].
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