A) \[{{h}_{1}}\,=\,\,2{{h}_{2}}=3{{h}_{3}}\]
B) \[{{h}_{1}}\,=\,\,\frac{{{h}_{2}}}{3}\,\,=\,\,\frac{{{h}_{3}}}{5}\]
C) \[{{h}_{2}}\,=\,\,3{{h}_{1}}\,and\,\,{{h}_{3}}=3{{h}_{2}}\]
D) \[{{h}_{1}}\,\,=\,\,{{h}_{2}}\,=\,{{h}_{3}}\]
Correct Answer: B
Solution :
For a particle released from a certain height the distance covered by the particle in relation with time is given by, \[h\,\,=\,\,\frac{1}{2}\,g{{t}^{2}}\] For first 5 sec, \[{{h}_{1}}=\,\,\frac{1}{2}\,g{{(5)}^{2}}=125\] Further next 5 sec, \[{{h}_{1}}+{{h}_{2}}=\,\,\frac{1}{2}\,g{{(10)}^{2}}=500\] \[\Rightarrow \,\,\,\,{{h}_{2}}=375\] \[{{h}_{1}}+{{h}_{2}}+{{h}_{3}}=\,\,\frac{1}{2}\,g{{(15)}^{2}}=\,\,1125\] \[\Rightarrow \,\,\,\,{{h}_{3}}=\,\,625\] \[{{h}_{1}}=3{{h}_{1}},\,\,{{h}_{3}}=5{{h}_{1}}\] \[or\,\,\,\,\,{{h}_{1}}=\frac{{{h}_{2}}}{3}\,\,=\,\,\frac{{{h}_{3}}}{5}\]
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