A) 6 m
B) \[\frac{2}{3}\] m
C) \[\frac{2}{9}\] m
D) 18 m
Correct Answer: D
Solution :
It is given that, acceleration due to gravity on planet A is 9 times the acceleration due to gravity on planet B i.e., \[{{g}_{A}}=9{{g}_{a}}\] (i) From third equation of motion \[{{v}^{2}}=2gh\] At planet A, \[{{h}_{A}}=\frac{{{v}^{2}}}{2{{g}_{A}}}\] (ii) At planet B, \[{{h}_{B}}=\frac{{{v}^{2}}}{2{{g}_{B}}}\] (iii) Dividing Eq. (ii) by Eq. (iii), we have \[\frac{{{h}_{A}}}{{{h}_{B}}}=\frac{{{g}_{B}}}{{{g}_{A}}}\] From Eq. (i), \[{{g}_{A}}=9{{g}_{B}}\] \[\therefore \frac{{{h}_{A}}}{{{h}_{B}}}=\frac{{{g}_{B}}}{9{{g}_{B}}}=\frac{1}{9}\] or \[{{h}_{B}}=9{{h}_{A}}=9\times 2=18\,m(\because \,\,{{h}_{A}}=2m)\]
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