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Sagot :
Answer:
22.18 m/s
Explanation:
We will use the following equation:
[tex]v^2_f_{}=v^2_i+2ay[/tex]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:
[tex]\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}[/tex]So, the ball strikes the ground at 22.18 m/s
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