Question

a man throws ball with the same speed vertically upward 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 the same time given G is equals to 9.8 metre per second square

Answer 1

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

Thus, the ball must be thrown at speed more than or equal to 19.6 m/s

Answer 2

atleast 19.6 m/s........to be 3 balls together in sky..... time interval is 2+2= 4 S.... so first ball must remain atleast 4s in sky..... so t=4/2=2s ....v=gt=19.6m/s