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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, $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
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(Assume, $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$5 \mathrm{~m}$
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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