Answer:
t = 3.38 s
Step-by-step explanation:
We have,
Initial speed of the ball that leaves the student's hand is 16 m/s
Initially, the hand is 1.90 m above the ground.
It is required to find the time for which the ball in the air before it hits the ground. We can use the equation of kinematics as :

Here,
and a=-g
The equation become:

After rearranging we get the above equation as :

It is a quadratic equation, we need to find the value of t. On solving the above equation, we get :
t = -0.115 s and t = 3.38 s (ignore t = -0.115 s )
So, the ball is in air for 3.38 seconds before it hits the ground.