Answer
(a) $9.29\frac{m}{s}$
(b) $56.4^{\circ}$ below horizontal.
Work Step by Step
(a) First of all, we convert the given initial speed into SI units as follows:
$v_{\circ x}=10\frac{naut.mi}{hr}\times \frac{1852m}{naut \space mi}\times \frac{1hr}{3600s}=5.14\frac{m}{s}$
We know that
$v_y^2=v_{\circ y}^2-2g\Delta y$
$\implies v_y=\pm \sqrt{(0)^2-2(9.81)(-10.0ft\times 0.305m/ft)}$
$v_y=-7.74\frac{m}{s}$
Now we can find the final speed as
$v=\sqrt{v_x^2+v_y^2}$
We plug in the known values to obtain:
$v=\sqrt{(5.14)^2+(-7.7)^2}$
$v=9.29\frac{m}{s}$
(b) We can find the direction of motion as
$\theta=tan^{-1}(\frac{v_y}{v_x})$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{-7.74}{5.14})$
$\theta=-56.4^{\circ}$
The negative sign shows that it is below horizontal.
