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

• A

  15 m
• Correct answer
  B

  10 m
• C

  20 m
• D

  5 m

Solution

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 $v^2=u^2-2 g h$ At $A$, $(10)^2=u^2-2 \times 10 \times \frac{H}{2}$ $\Rightarrow u^2=100+10 H...(i)$ At $O^{\prime},(0)^2=u^2-2 \times 10 \times H$ $\Rightarrow u^2=20 H...(ii)$ Thus, from Eqs. (i) and (ii), we get $20 H=100+10 H$ $\Rightarrow 10 H=100 \quad \therefore H=10 \mathrm{~m}$

Hence, the correct answer is option: 'B'.