Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A man throws balls with the same speed vertically upwards one after the other at an interval of $2 \text{ seconds}$ . What should be the speed of the throw so that more than two balls are in the sky at any time? (Given $g=9.8 \text{ m/s}^2$ )

At least $0.8 \text{ m/s}$

Any speed less than $19.6 \text{ m/s}$

Only with speed $19.6 \text{ m/s}$

More than $19.6 \text{ m/s}$

Step-by-Step Solution

Analyze the Condition: The problem states that balls are thrown at intervals of $\Delta t = 2 \text{ s}$ . For 'more than two balls' (i.e., at least 3) to be in the sky simultaneously, the first ball must remain in the air until the third ball is thrown.
Time Constraint: Let the time of flight for a single ball be $T$ . If the third ball is thrown at $t = 2\Delta t = 4 \text{ s}$ , the first ball (thrown at $t=0$ ) must not have landed yet. Therefore, the time of flight must be greater than $4 \text{ s}$ ( $T > 4 \text{ s}$ ).
Formula for Time of Flight: For an object thrown vertically upwards with speed $u$ and returning to the same level, the total time of flight is given by $T = \frac{2u}{g}$ .
Calculation: $\frac{2u}{g} > 4$ $2u > 4g$ $u > 2g$ Substitute $g = 9.8 \text{ m/s}^2$ : $u > 2 \times 9.8$ $u > 19.6 \text{ m/s}$
Conclusion: The speed of the throw must be more than $19.6 \text{ m/s}$ .

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