Answer:
22.18 m/s
Step-by-step explanation:
We will use the following equation:

Where vf is the final velocity
vi is the initial velocity, so it is -10 m/s
a is gravity, so it is -9.8 m/s²
y is the change in the height so it is -20 m
Therefore, replacing the values, we get:
![\begin{gathered} v^2_f=(-10)^2+2(-9.8)(-20) \\ v^2_f=100+392 \\ v^2_f=492 \\ v_f=\sqrt[]{492}=22.18\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/oupot1m7nrrgoleeybyu8x0ny9iqljc1z9.png)
So, the ball strikes the ground at 22.18 m/s