A) 15 m
B) 10 m
C) 20 m
D) 5 m
Correct Answer: B
Solution :
Let maximum height attained by the ball be H. Using \[{{v}^{2}}={{u}^{2}}-2gh\] \[At\,\,A,\,\,{{(10)}^{2}}\,=\,\,{{u}^{2}}-2\times 10\times \frac{H}{2}\] \[\Rightarrow \,\,\,{{u}^{2}}=100\,\,+\,\,10\,H\] ? (i) \[At\text{ }O',\text{ }{{(0)}^{2}}={{u}^{2}}-2\times 10\times H\] \[\Rightarrow \,\,\,{{u}^{2}}= 20\,\,H\] Thus, from Eqs. (i) and (ii), we get \[20\,\,H=100+10\,\text{H}\] \[\Rightarrow \,\,\,10\,\,H=100\,\,\,\,\,\,\therefore \,\,\,H=10\,m\]You need to login to perform this action.
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