Equation of motion:
\(y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}\)
Since initial velocity is zero, the second term goes away:
\(y_{f}=y_{o}+0+ \frac{1}{2} at^{2}\)
\(y_{f}=y_{o}+\frac{1}{2} at^{2}\)
\(y_{f}-y_{o}= \frac{1}{2} at^{2}\)
\(y_{f}-y_{o}=5.4m\)
\(5.4m= \frac{1}{2} at^{2}\)
\(\frac{2(5.4m) }{a} = t^{2}\)
\(a = g = 9.81 \frac{m}{ s^{2}}\)
\(\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}\)
\(1.1 s^{2} = t^{2}\)\(\sqrt{1.1 s^{2}} = \sqrt{t^{2}}\)
\(t = 1.05s\)