Answer
Final Speed at the top is 2.5 m/s
Pressure at the top is 2.2 atm
Work Step by Step
Solving for speed at the top:
In this, subscript 1 means initial or at the bottom and subscript 2 means final or at the top. Therefore:
$A_1v_1 = A_2v_2$
Solving for $v_2$
$v_2 = \frac{A_1v_1}{A_2} = \frac{.025^2*3.14*.78}{.014^2*3.14}$ = 2.5 m/s
For the Pressure, we use the Bernoulli's Principle:
$P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2$
Because $y_1$ = 0, that term is gone so we can solve for $P_2$ by saying:
$P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2 - \rho g y_2$
With the density of water being 1000
= (3.8*101300) + ($\frac{1}{2}$ * 1000*.78$^2$) - ($\frac{1}{2}$ * 1000*2.5$^2$) - (1000*9.8*16)
=225319.2/101300 = 2.2 atm
