Answer
The cross-sectional area of the stream is equal to one-half the area of the hole when the water is $~~0.90~m~~$ below the bottom of the tank.
Work Step by Step
We can find the speed when the cross-sectional area of the stream is equal to one-half the area of the hole:
$v_2~A_2 = v_1~A_1$
$v_2~\frac{A_1}{2} = \sqrt{2gh}~A_1$
$v_2 = \sqrt{8gh}$
$v_2 = \sqrt{(8)(9.8~m/s^2)(0.30~m)}$
$v_2 = 4.85~m/s$
We can find the distance below the hole where the water has this speed:
$v_f^2 = v_0^2+2ad$
$d = \frac{v_f^2 - v_0^2}{2a}$
$d = \frac{v_f^2 - 2gh}{2a}$
$d = \frac{(4.85~m/s)^2 - (2)(9.8~m/s^2)(0.30~m)}{(2)(9.8~m/s^2)}$
$d = 0.90~m$
The cross-sectional area of the stream is equal to one-half the area of the hole when the water is $~~0.90~m~~$ below the bottom of the tank.
