A) 15m
B) 10m
C) 20m
D) 5m
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 \[20H=100+10H\] \[\Rightarrow \] \[10H=100\,\] \[\therefore \]\[H=10\,m\]You need to login to perform this action.
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