A) \[5m\]
B) \[150m\]
C) \[20m\]
D) \[10m\]
Correct Answer: D
Solution :
Let \[u\] be the initial velocity and \[h\] be the maximum height attained by the stone. \[v_{1}^{2}={{u}^{2}}-2g{{h}_{1}}\] \[\therefore \]\[{{(10)}^{2}}={{u}^{2}}-2\times 10\frac{h}{2}\left( \because {{h}_{1}}=\frac{h}{2},\,\,{{v}_{1}}=10m{{s}^{-1}} \right)\] or\[100={{u}^{2}}-10h\] ... (i) Again at height\[h\], \[v_{2}^{2}={{u}^{2}}-2gh\] \[0={{u}^{2}}-2\times 10\times h\] \[(\because \,\,{{v}_{2}}=0)\] \[0={{u}^{2}}-20h\] ... (ii) Subtracting Eq. (ii) from Eq. (i), we get \[100=10h\] or \[h=\frac{100}{10}=10m\]
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