A) 3 m
B) 15 m
C) 1 m
D) 10 m
Correct Answer: D
Solution :
Let u be the initial velocity and h be the maximum height attained, by the stone \[{{v}_{1}}^{2}={{u}^{2}}-2gh,\] \[{{(10)}^{2}}={{u}^{2}}-2\times 10\times \frac{h}{2}\] \[100={{u}^{2}}-10h\] ?(i) Again at height h \[{{v}_{2}}^{2}={{u}^{2}}-2gh\] \[{{(0)}^{2}}={{u}^{2}}-2\times 10\times h\] \[{{u}^{2}}=20h\] So, from Eqs. (i) and (ii), we have \[100=10h\] \[h=10\,m\]You need to login to perform this action.
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