A) 15 m
B) 10 m
C) 20 m
D) 5 m
Correct Answer: B
Solution :
Key Idea: The problem can be solved using third equation of motion at A and O. |
Let maximum height attained by the ball be H. third equation of motion gives |
\[{{v}^{2}}={{u}^{2}}-2gh\] |
At, \[A,\] \[\,{{(10)}^{2}}={{u}^{2}}-2\times 10\times \frac{H}{2}\] |
\[\Rightarrow \] \[{{u}^{2}}=100+10\,H\] (i) |
At \[O',\] \[{{(0)}^{2}}={{u}^{2}}-2\times 10\times H\] |
\[\Rightarrow \] \[{{u}^{2}}=20\,H\] (ii) |
Thus, from Eqs. (i) and (ii), we get |
\[20H=100+10H\] |
\[\Rightarrow \] \[10H=100\] |
\[\therefore \] \[H=10\,m\] |
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