In order to find the flight time, use the following formula:

where,
y: final height = 0 m
yo: initial height = 30 m
g: gravitational acceleration constant = 9.8 m/s^2
t: time
Due to vo = 0m/s and y = 0 m, you can solve the equation above for t, as follow:

Then, by replacing the values of the parameters, you get:
![t=\sqrt[]{(2(30m))/((9.8m)/(s^2))}=2.47s](https://img.qammunity.org/2023/formulas/physics/college/t5395tvmlg460rzdahojqvthzlc1ht6a1w.png)
Hence, the ball takes 2.74 s to hit the ground
To find the velocity just before the ball hits the ground, use the following formula:

Replace the values of the parameters:

Hence, the speed of the ball is approximately 24.2 m/s