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 \[g=10m/{{s}^{2}}\])          [AIPMT (S) 2005]

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

\[20H=100+10H\]

\[\Rightarrow \]   \[10H=100\]

\[\therefore \]      \[H=10\,m\]