First, we'll find h(2):
[tex]
h(t)=-4.9t^2+20\\\\
h(2)=-4.9\cdot2^2+20\\\\
h(2)=-4.9\cdot4+20\\\\
h(2)=-19.6+20\\\\
\boxed{h(2)=0.4~m}
[/tex]
Now, we'll find when h(t)=0:
[tex]
h(t)=0\\\\
-4.9t^2+20=0\\\\
4.9t^2=20\\\\
t^2=\dfrac{20}{4.9}\\\\
t^2\approx4.08\\\\
t\approx\pm\sqrt{4.08}\\\\
\to\text{Since t represents the elapsed time, it is always positive:}\\\\
t\approx\sqrt{4.08}\\\\
\boxed{t\approx2.02~s}
[/tex]
