Answer
The ball will be in the air for a total of $6.4s$.
Work Step by Step
Step-1: From the previous question, $v_0=31ms^{-1}$. For an ascending ball, $$v=v_0-at$$$$0=31-9.8t$$$$\implies t=\frac{31}{9.8}=3.16s \approx 3.2s$$Thus, the ball takes $3.2s$ to go up to a maximum height of $50m$.
Step-2: The ball will take the same time to come down. Thus, the ball remains in the air for a total time of $t=2\times 3.2=6.4s$.
