Answer
Please see the work below.
Work Step by Step
We know that
$t=\sqrt{\frac{2h}{g}}$
We plug in the know values to obtain:
$t=\sqrt{\frac{(2)(4.2-1.5)}{9.8}}=0.742s$
The horizontal speed is
$v_x=\frac{x}{t}=\frac{3.0}{0.742}=4.04\frac{m}{s}$
The vertical speed is
$v_y=gt=9.8(0.742)=7.27\frac{m}{s}$
Now we can find the speed of the package as
$v=\sqrt{(v_x)^2+(v_y)^2}=\sqrt{(4.04)^2+(7.27)^2}=8.3\frac{m}{s}$
and the direction of the package is
$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{7.27}{4.04})=61^{\circ}$
