Physics-
General
Easy

Question

A man throws balls with the same speed vertically upwards one after the other at an interval of two seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given g equals 9.8 blank m divided by s to the power of 2 end exponent)

  1. At least 0.8 blank m divided by s  
  2. Any speed less than 19.6 blank m divided by s  
  3. Only with speed 19.6 blank m divided by s  
  4. More than 19.6  

The correct answer is: More than 19.6


    Interval of ball throw equals 2 s e c
    If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than 4 blank s e c
    T greater than 4 blank s e c
    fraction numerator 2 u over denominator g end fraction greater than 4 sec invisible function application rightwards double arrow u greater than 19.6 blank m divided by s
    For u equals 19.6, first ball will just about to strike the ground (in air)
    Second ball will be at highest point (in air)
    Third ball will be at point of projection or at ground (not in air)

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