check the picture below. So it hits the ground when y = 0.
[tex]\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{\textit{0, because the hammer was at rest}}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{\textit{20 ft, so he was 20 ft above ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases}[/tex]
[tex]\bf h(t)=\stackrel{\textit{from gravity}}{\stackrel{\downarrow }{-16t^2}+0t+20}\implies h(t)=-16t^2+20 \\\\\\ \stackrel{\textit{we set h(t) = 0}}{0=-16t^2+20}\implies 16t^2=20\implies t^2=\cfrac{20}{16}\implies t^2=\cfrac{5}{4} \\\\\\ t=\sqrt{\cfrac{5}{4}}\implies t=\cfrac{\sqrt{5}}{\sqrt{4}}\implies t=\cfrac{\sqrt{5}}{2}\implies t\approx 1.12[/tex]
