Answer:
...
<h2>PE=
<em>work done</em></h2><h2>
<em>m</em><em>gh</em><em>=</em><em>2</em><em>0</em><em>×</em><em>1</em><em>0</em><em>×</em><em>2</em><em>0</em><em>.</em><em>.</em></h2>

.
<em>I </em><em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
12.5m/s
Explanation:
(Assuming the question was asking for the speed just before it hit the ground)
We can use the first key equation of accelerated motion
Vf^2 = Vi^2+2aΔd
Vf^2 = 0 + 2(9.8)(8) (plugged in values, initial velocity is 0 since the ball was at rest)
Vf^2 = 156.8
Vf = 12.5 (squared both sides)