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? (Take \[g=10\,\,m/{{s}^{2}}\])

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