Answer
$28.2\;\rm m/s$
Work Step by Step
We need to package to reach the explorer position 200-m below the package and 30-m far from the bottom of the cliff, as shown in the given figure.
We know that the package will be fired horizontally, so its initial speed will be $v_0=v_{ix}$ and hence, $v_{iy}=0$. This means that the package will be under free-fall acceleration.
Hence, we need to find the time it takes to reach the ground.
$$y=y_i+v_{iy}t+\frac{1}{2}a_yt^2$$
where $y=0$ m, $y_i=200$ m, $v_{iy}=0$, and $a_y=-g$
$$0=y_i+0-\frac{1}{2}gt^2$$
solving for $t$;
$$t=\sqrt{\dfrac{2y_i}{g}}=\sqrt{\dfrac{2\times 200}{9.8}}=\bf 6.39\;\rm s$$
We assume that air resistance is negligible, so the horizontal velocity component remains constant during its trip.
Thus,
$$v_{ix}=v_0=\dfrac{\Delta x}{ t}$$
Plugging the known;
$$ v_0=\dfrac{30}{ 6.39}=\bf 4.7\;\rm m/s$$
Now we can use the momentum principle; we assume that the rocket and the package are an isolated system during the collision.
Thus,
$$p_{ix}=p_{fx}$$
We know that the rocket and the package will move after the collision as one unit.
$$m_Rv_{ix,R}+m_Pv_{ix,P}=(m_R+m_P)v_0$$
and we know that the package was initially at rest, so
$$m_Rv_{ix,R}+0=(m_R+m_P)v_0$$
Solving for $v_{ix,R}$
$$v_{ix,R} =\dfrac{(m_R+m_P)v_0 }{m_R}$$
Plugging the known;
$$v_{ix,R} =\dfrac{(1+5)4.7}{1}=\color{red}{\bf 28.2}\;\rm m/s$$
