Answer
$18\frac{m}{s}$
Work Step by Step
We can find the required speed as follows:
$v_y^2=v_{\circ}^2sin^2\theta-2g\Delta y$
This can be rearranged as:
$v_y=\sqrt{v_{\circ}^2sin^2\theta-2g\Delta y}$
We plug in the known values to obtain:
$v_y=\sqrt{(12)^2sin^2(-40^{\circ})-2(9.81)(-10)}$
$v_y=18\frac{m}{s}$
