A person throws balls upward in regular intervals of time of 2 s. What should be the speed of throwing of balls such that both balls remain in air ?

A person throws balls upward in regular intervals of time of 2 s. What

1.	A person throws balls upward in regular intervals of time of 2 s. What should be the speed of throwing of balls such that both balls remain in air ?
Answer» less than `19.6 ms^(-1)` EQUAL to `19.6 ms ^(-1)` less than `9.8 ms^(-1)` greater than `19.6 ms^(-1)` Solution :Displacement covered for time of flight (T), From equation `THEREFORE h = v _(0) t + 1/2 at ^(2)` `therefore 0= v _(0) T -1.2 G T ^(2)` `therefore T= (2v _(0))/(g) ""…(1)` Regular intervals are of 2s. Suppose minimum three (means more than two) balls remain in air, for that time of flight of first ball should be greater than 4s, `therefore (2v _(0))/(g) gt 4` `therefore v _(0) gt (49)/(2)` `therefore v _(0) gt (4 xx 9.8)/(2)` `therefore v _(0) gt 19.6 ms^(-1)` THus, if `v _(0) gt 19.6 ms ^(-1)` so when the first ball is about to come in hand at that time second ball will be at MAXIMUM height and third ball will be about to be released.