Answer
$v_0=\frac{1}{2t_0}(2h-gt_0^2)$
Work Step by Step
Let the path of object is represented by $y(t)$ which moves under the effect of gravity then,
$\frac{d^2y}{dt^2}=g$
where $g$ repersents the acceleration due to gravity
Also, $y(0)=0$, $y(t_0)=h$, $\frac{dy}{dt}(0)=v_0$
Integrate the equation $\frac{d^2y}{dt^2}=g$,
$\frac{dy}{dt}=gt+c_1$
Using intial condition $\frac{dy}{dt}(0)=v_0$
$v_0=g(0)+c_1$
$c_1=v_0$
On putting value of $c_1$ we have,
$\frac{dy}{dt}=gt+v_0$
on integrating again
$y=\frac{1}{2}gt^2+v_0t+c_2$
Using intial condition $y(0)=0$ we have,
$0=\frac{1}{2}g(0)^2+v_0(0)+c_2$
$c_2=0$
Now equation becomes $y=\frac{1}{2}gt^2+v_0t$
Use intial valve $y(t_0)=h$
$h=\frac{1}{2}gt_0^2+v_0\cdot t_0$
$h-\frac{1}{2}gt_o^2=v_0\cdot t_0$
$\frac{2h-gt_0^2}{2}=v_0\cdot t_0$
$v_0=\frac{1}{2t_o}(2h-gt_0^2)$
