KCET · Physics · Motion In One Dimension
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 \(20 \mathrm{~m}\)
- B \(5 \mathrm{~m}\)
- C \(10 \mathrm{~m}\)
- D \(7.5 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The height to which the balls
\(\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \times 10 \times 1^{2}=5 \mathrm{~m}\)
\(\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \times 10 \times 1^{2}=5 \mathrm{~m}\)
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