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A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time?
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The correct answer is:
more than $19.6 \mathrm{~m} / \mathrm{s}$
From question Interval of ball thrown $=2 \mathrm{sec}$
If we want that minimum three balls remain in air then time of flight of first ball must be greater than 4 sec.
$\begin{aligned}
& T>4 \mathrm{sec} \\
& \therefore \frac{2 u}{g}>4 \mathrm{sec} \\
& \Rightarrow \quad u>19.6 \mathrm{~m} / \mathrm{s}
\end{aligned}$
If we want that minimum three balls remain in air then time of flight of first ball must be greater than 4 sec.
$\begin{aligned}
& T>4 \mathrm{sec} \\
& \therefore \frac{2 u}{g}>4 \mathrm{sec} \\
& \Rightarrow \quad u>19.6 \mathrm{~m} / \mathrm{s}
\end{aligned}$
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