Q175 / 180 · KCET 2012 · Physics
A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is
(Assume, $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
- A$20 \mathrm{~m}$
- Correct answerB$5 \mathrm{~m}$
- C$10 \mathrm{~m}$
- D$7.5 \mathrm{~m}$
Solution
The height to which the balls
$$
\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \times 10 \times 1^{2}=5 \mathrm{~m}
$$
Hence, the correct answer is option: 'B'.