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 ?

Options

A. less than `19.6 ms^(-1)`
B. equal to `19.6 ms ^(-1)`
C. less than `9.8 ms^(-1)`
D. greater than `19.6 ms^(-1)`

Answer

D

Explanation

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.