A) 5 m
B) 150 m
C) 20 m
D) 10 m
Correct Answer: D
Solution :
Let u be the initial velocity and h be the maximum height attained by the stone. \[v_{1}^{2}={{u}^{2}}-2g{{h}_{1}}\] \[\therefore {{(10)}^{2}}={{u}^{2}}-2\times 10\frac{h}{2}\,\left( \because {{h}_{1}}=\frac{h}{2},\,{{v}_{1}}=10\,m{{s}^{-1}} \right)\] or \[100={{u}^{2}}-10\,h\] ?(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 \[100=10\,h\] or \[h=\frac{100}{10}=10\,m\]You need to login to perform this action.
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