A man continuously throws balls into the sky vertically within 2 sec of time interval. At what speed he should throw the balls so that more than two balls should be in the sky at any time. (Take g = 10 m/s2)

Interval of ball throw = 2 sec If we want that minimum three (more than two) balls remain in air then time of flight of first ball must be greater than 4 sec T > 4 secFrom v = u - gt ; v = 0, time for half the distance travelled d=ugThus for the total distance the time 2ug>4sec⇒u>20m/sFor u = 20 m/s, first ball will just strike the ground (in sky) second ball will be at highest point (in sky) and third ball will be at point of projection or at ground (not in sky)