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 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 m/s

Any speed less than 19.6 m/s

Only with speed 19.6 m/s

More than 19.6 m/s

Step-by-Step Solution

Analyze the Condition: For more than two balls (i.e., at least 3 balls) to be in the sky simultaneously, the first ball must remain in the air when the third ball is thrown.
Time Intervals: Ball 1 is thrown at $t=0$ . Ball 2 is thrown at $t=2\text{ s}$ .

Ball 3 is thrown at $t=4\text{ s}$ .

Apply Kinematics: The time of flight ( $T$ ) of the first ball must be greater than 4 seconds so that it is still in the air when the third ball is projected at $t=4\text{ s}$ .

Formula for time of flight: $T = \frac{2u}{g}$ .

Solve for Speed ( $u$ ): $T > 4$ $\frac{2u}{g} > 4$ $u > 2g$ $u > 2 \times 9.8$ $u > 19.6 \text{ m/s}$

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