Answer
$348.7ft/s$, $237.8 mi/hr$
Work Step by Step
Step 1. We can model the height of the lava with a function $s(t)=v_0t-16t^2$ where $v_0$ is the lava's exit velocity (ft/s).
Step 2. Taking the derivative, we get $v(t)=s'(t)=v_0-32t$ and when the lava reaches its maximum height $v(t)=0$, we get $t=v_0/32$
Step 3. With the record height $s=1900$ ft, we have $1900=v_0(v_0/32)-16(v_0/32)^2$ or $\frac{v_0^2}{64}=1900$, thus we can obtain $v_0=\sqrt {1900\times64}=80\sqrt {19}\approx348.7ft/s=348.7ft/s\times\frac{3600s/hr}{5280ft/mi}\approx237.8 mi/hr$
