Answer
$150m/s$
Work Step by Step
$F_B=mg=(1.7\times10^6kg)(9.81m/s^2)=1.67\times10^7N$
$P_2+\frac{1}{2}\rho v_2^2+\rho gy_2=P_1+\frac{1}{2}\rho v_1^2+\rho gy_1$
$\rho=1.29 kg/m^3$; $v_2=95m/s$; $y_1=y_2$
$P_1=\frac{F_{Lift}}{A}$
$P_2=\frac{W}{A}$
$\frac{W}{A}=\frac{1}{2}\rho (v_1^2-v_2^2)$
$v_1=\sqrt{2\frac{W}{A\rho}+v_2}=\sqrt{2\frac{1.67\times10^7N}{(1200m^2)(1.29kg/m^3)}+95m/s}=150m/s$
