Q103 / 200 · NEET 2003 · Physics

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?

  • Correct answer
    A
    more than $19.6 \mathrm{~m} / \mathrm{s}$
  • B
    at least $9.8 \mathrm{~m} / \mathrm{s}$
  • C
    any speed less than $19.6 \mathrm{~m} / \mathrm{s}$
  • D
    only with speed $19.6 \mathrm{~m} / \mathrm{s}$

Solution

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. $\begin{aligned} & T>4 \mathrm{sec} \\ & \therefore \frac{2 u}{g}>4 \mathrm{sec} \\ & \Rightarrow \quad u>19.6 \mathrm{~m} / \mathrm{s} \end{aligned}$

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

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