Question
A man throws balls with same speed vertically upwards one after the other at an interval of $$2\,\sec .$$ What should be the speed of throw so that more than two balls are in air at anytime?
A.
Only with speed $$19.6\,m/s$$
B.
More than $$19.6\,m/s$$
C.
At least $$9.8\,m/s$$
D.
Any speed less then $$19.6\,m/s.$$
Answer :
More than $$19.6\,m/s$$
Solution :
Height attained by balls in $$2\,\sec $$ is $$ = \frac{1}{2} \times 9.8 \times 4 = 19.6\,m$$
the same distance will be covered in 2 second (for descent)
Time interval of throwing balls, remaining same. So, for two balls remaining in air, the time of ascent or descent must be greater than 2 seconds. Hence speed of balls must be greater than $$19.6\,m/\sec .$$