Each ball has acceleration -g= -9.81 m/s^{2}. If we take "v" to be the initial velocity of each ball, then v(T)= v- 9.81T in m/s where T is measured in seconds since the ball was thrown.
In addition, the height of each ball after T seconds is given by
h(T)= vT- (9.81/2)T^{2}.
The ball will continue upward until its velocity is 0: v- 9.81T= 0 so
T= v/9.81. We are told that the height at that time is 3 m:
v(v/9.81)- (9.81/2)(v/9.81)^{2}= v^{2}/(2*9.81)= 3 so
v^{2}= 2*3*9.81 and v= &radi;(2*3*9.81).

You can find t in either of two ways: directly- The ball comes back to the person's had (assumed same height) exactly t seconds after the sixth ball is throne up and there is and interval of t seconds between each one: the first ball comes back to the person's had after 6t seconds: h(6t)= v(6t)- (9.81/2)(6t)^{2}= 0. You know v from the previous problem. Solve for t.
Less directly: because of the symmetry of the motion, each ball attains it highest point exactly half way through its motion: 3t.
Since you calculated that time, T, in the first problem, 3t= T. Solve for t.

You know that when the first ball is again in the juggler's hand, time 6t has passed. Its height is0. The second ball is exactly "t" behind. Its height will be given by h(5t), the third ball's height will be h(4t), the fourth ball's height by h(3t), the fifth ball's height by h(2t), and the sixth ball's height by h(t). If the juggler throws the first ball up again the instant he catches it, each ball will have one of those 6 heights each time a ball is caught.