Answer
$9.33\times10^{-5}m^{2}$
Work Step by Step
Let's apply the equation $V^{2}=u^{2}+2aS$ in the vertical direction to find the speed of the water below the faucet.
$\downarrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$V^{2}=(0.85\space m/s)^{2}+2(9.8\space m/s^{2})(0.1\space m)$
$V=1.64\space m/s$
Now we apply the equation $A_{1}v_{1}=A_{2}v_{2}$ to find the area of the stream at a distance 01 m below the faucet.
$A_{1}v_{1}=A_{2}v_{2}$ ; Let's plug known values into this equation.
$(1.8\times10^{-4}m^{2})(0.85\space m/s)=A_{2}(1.64\space m/s)$
$A_{2}=9.33\times10^{-5}m^{2}$
