Note: Since I could not understand the height it was dropped from I will be using R. You can substitute whatever height in place of R.
Let the up be increasing y and down be decreasing y. We know the acceleration in the y direction which is,
[tex]\ddot{y}=-10\frac{m}{s^{2}}=-g[/tex]
I'm going to call it -g for simplicity.
Now integrating to get velocity,
[tex]\dot{y}=-gt[/tex]
Notice since the ball was dropped from rest, so the constant of integration is 0.
Integrating once more to get position,
[tex]\[y=R-\frac{1}{2}gt^{2}\][/tex]
<Since R is the initial height, it is the constant of integration here>
To answer the question for what time will it hit the ground, setting y=0 and solving for t we get:
[tex]\[t=\sqrt{2gR}\][/tex]
Now to get the velocity at which it hit the ground, we plug in this value of t into our velocity equation:
[tex]\dot{y}(t=\sqrt{2gR}\)=-g\sqrt{2gR}[/tex]
Hope this helps!
