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Dear chakrala mounica
the first has to be above the ground when the second ball is thrown from the ground
so the first ball has to be above from the ground for 2 seonds.
s=ut-1/2*gt^2 t=2 and s=0 when first ball reaches the ground
0= 2*u- 1/2*g*4
u=g m/s take g=9.8 m/s^2
u= 9.8 m/s
speed has to be more than 9.8 m/s
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Askiitians Expert
Rathod Shankar Singh
IIT Bombay
For at least 3 balls to remain above the ground, the time of flight of the first ball must be more than or equal to 4 s. Because in t = 4 s the third ball leaves the ground.
Now, for the first ball,
Initial velocity = u
Final velocity = -u
Acceleration = -g
So, -u = u – gt
=> u = gt/2
=> u = 9.8 × 4/2
=> u = 19.6 m/s
so the speed should be greater than 19.6 m/s to keep more than 2 balls in the sky at any time.
Thus, the ball must be thrown at speed more than or equal to 19.6 m/s u
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