Answer
4.49 Mm/s at 43 degrees below the horizontal.
Work Step by Step
Using conservation of momentum, we find the x and y components of the velocity to be as follows:
$v_{px}=\frac{(5)(1.6)-4cos(33^{\circ})(1.4)}{1}=3.3$
$v_{py}=\frac{(1.4)(4)(sin33^{\circ})}{1}=-3.05$
We find the speed to be:
$v=\sqrt{3.3^3+(-3.05)^2}=4.49 \ Mm/s$
We find the angle:
$\theta = tan^{-1}(-3.05/3.3)=-43^{\circ}$
