question_answer

                A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height. How high does the ball rise? (Taking \[g=10\,m/{{s}^{2}}\])                         

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                 \[20H=100+10H\]                         \[\Rightarrow \]          \[10H=100\] \[\therefore \] \[H=10\,m\]