Answer
The object's speed as it hits the ground is $\sqrt{\frac{2~G~M~h}{R^2+R~h}}$
Work Step by Step
We can use conservation of energy to find the object's speed as it hits the ground.
$K_f+U_f = K_0+U_0$
$K_f = 0+U_0-U_f$
$\frac{1}{2}mv^2 = -\frac{G~M~m}{R+h}-(-\frac{G~M~m}{R})$
$\frac{1}{2}mv^2 = \frac{G~M~m}{R}-\frac{G~M~m}{R+h}$
$\frac{1}{2}v^2 = \frac{(G~M)(R+h)}{(R)(R+h)}-\frac{G~M~R}{(R)(R+h)}$
$v^2 = \frac{2~G~M~h}{R^2+R~h}$
$v = \sqrt{\frac{2~G~M~h}{R^2+R~h}}$
The object's speed as it hits the ground is $\sqrt{\frac{2~G~M~h}{R^2+R~h}}$
