#### Answer

$F=303lb$

#### Work Step by Step

The required force can be determined as follows:
$m=\frac{\gamma}{g}\times Q=\frac{62.4}{32.2}\times 2=3.876slug/s$
Similarly $v_{Ay}=\frac{2}{0.02778}=72ft/s$
and the velocity of the water in the horizontal direction is
$v_{Bx}=\frac{2}{0.08333}=24ft/s$
We apply the impulse and momentum principle in the vertical direction
$F_y=\frac{dm}{dt}(v_{Ay}-v_{By})$
$\implies F_y=3.876(72-0)=279.072lb$
Now we apply the impulse and momentum principle in the horizontal direction
$\Sigma F_x=\frac{dm}{dt}(v_{Ax}-v_{Bx})$
We plug in the known values to obtain:
$24-F_x=3.8760(0-24)$
$\implies F_x=117.01lb$
The required resultant force can be calculated as
$F=\sqrt{F_x^2+F_y^2}$
We plug in the known values to obtain:
$F=\sqrt{(117.01)^2+(279.02)^2}$
This simplifies to:
$F=303lb$