Answer
$A\sqrt{\frac{2g(h_1-h_2)}{3}}$
Work Step by Step
We know that according to Bernoulli's principle
$P_1+\frac{1}{2}\rho V_1^2+\rho gh_1=P_2+\frac{1}{2}\rho V_2^2$...............eq(1)
We also know that
$P_2=P_{atm}+\rho gh_2$
According to equation of continuity
$A_1V_1=A_2V_2$
$\implies AV_1=\frac{A}{2}V_2$
so $V_2=2V_1$
We plug in the known values in equation(1) to obtain:
$P_{atm}+\frac{1}{2}\rho V_1^2+\rho gh_1=P_{atm}+\rho gh_2+\frac{1}{2}\rho (2V_1)^2$
This simplifies to:
$\rho g(h_1-h_2)=\frac{1}{2}\rho (4V_1^2-V_1^2)$
$\implies \frac{2g(h_1-h_2)}{3}=V_1^2$
Thus $V=A_1V_1=A\sqrt{\frac{2g(h_1-h_2)}{3}}$
