Starting with the basic equations for the kinematics, we have object 1 falling from h with zero starting velocity, and object 2 going up from the ground: $$ x_1(t) = h - \frac{gt^{2}}{2} \tag{1.1} $$ $$ x_2(t) = v_f t + \frac{gt^{2}}{2} \tag{1.2} $$
Now, recall that for the falling body 1: $$ v = gt \tag{1.3}$$ We have to compute the final velocity, taking the final time, that is the time when the object 1 hit the ground. Using 1.1 for finding $t_f$, we have: $$ x_1(t_f) = 0 \Rightarrow t_f = \sqrt{\frac{2h}{g}} \tag{1.4}$$ Using 1.2 and inserting the final time 1.4 : $$ v_f = \sqrt{2gh} \tag{1.5} $$ And finally: $$ x_2(t) = \sqrt{2gh}t + \frac{gt^2}{2} \tag{1.6}$$ Now we have to write 1.1 and 1.2 taking respect to time. Finding the square root of 1.2 we have: $$ t_2 = \frac{\sqrt{2gh} + \sqrt{2gh - 2gx_2}}{g} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{2(h - x_2)}{g}} \tag{1.7}$$ And for 1.1: $$ t_1 = \sqrt{\frac{2(h - x_1)}{g}} \tag{1.8}$$ Since the two objects meet at some point in space and time, this two times are equal, $t_1 = t_2 $, and calling the position where they meet $ x = x_1 = x_2 $ we have: $$ \sqrt{\frac{2(h - x)}{g}} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{2(h - x)}{g}} \tag{1.9} $$ Taking the square: $$ \frac{2h}{g} - \frac{2x}{g} = \frac{2h}{g} + \frac{2(h - x)}{g} + \frac{4}{g}\sqrt{h(h - x)} \tag{1.10}$$ And then: $$ \frac{-2h}{g} = \frac{4}{g}\sqrt{h(h - x)} \tag{1.11}$$ $$ \frac{12h^2}{g^2} = \frac{16hx}{g^2} $$ In conclusion: $$ x = \frac{3}{4}h \tag{1.12}$$
Another method: we can use more simply the conservation of energy and the equations for the velocity: $$ v_1 = gt \tag{2.1} $$ $$ v_2 = v_f - gt \tag{2.2}$$ For finding $v_f$ we use the conservation of energy: $$ \frac{v_f^2}{2} = gh \tag{2.3}$$ Now using conservation of energy for object 1 falling: $$ gh = gx + \frac{v_1^2}{2} \tag{2.4}$$ For object 2 going up: $$ \frac{v_f}{2} = gx + \frac{v_2^2}{2} \tag{2.5} $$ Subtracting the latest two and replacing $v_f$, calling the common velocity v: $$ v := |v_2| = |v_1| \tag{2.6} $$ From 2.1 and 2.2 taking time and equating the two times when they meet, we find: $$ v_1 + v_2 = v_f \tag{2.7}$$ With 2.6 and 2.7 we find that $v = \sqrt{\frac{gh}{2}}$ If we insert this velocity in 2.4 or 2.5 we find: $$ x = \frac{3}{4}h \tag{2.8} $$ 771