Velocity of a rock

We have 
\[h''(t)=-g  \Rightarrow h'(t)=-gt+C,\]
where $C=h'(0)=v_0$, where $v_0$ is the initial velocity. So
\[(1)        h'(t)=-gt+v_0.\]

Integrating one more time we get 
\[h(t)=-\frac{1}{2}gt^2+v_0t+h_0.\]
Since we assume $0=h(0)=h_0$, we have $h_0=0$. Thus
\[(2)  h(t)=-\frac{1}{2}gt^2+v_0t.\]
When the rock reaches the maximum height $100$, it's speed is zero. Hence from (1) we get 
\[-gt+v_0=0 \Rightarrow t=\frac{v_0}{g}.\]
Substituting in (2) we get 
\[100=h(\frac{v_0}{g})=-\frac{1}{2}g(\frac{v_0}{g})^2+v_0(\frac{v_0}{g})=100.\]
Simplifying we get 
\[-\frac{1}{2}\frac{v^2_0}{g}+\frac{v^2_0}{g}=100  \Rightarrow \frac{1}{2}\frac{v^2_0}{g}=100\]
\[\Rightarrow v_0=\sqrt{200g}=44.27  m/s.\]