Answer
$\textbf{(a)}$ $v_h=2.70\text{ m/s}$.
$\textbf{(b)}$ $v_v=-20.6\text{ m/s}.$
$\textbf{(c)}$ The answer to part $\textbf{(a)}$ would increase.
The answer to part $\textbf{(b)}$ would remain the same.
Work Step by Step
$\textbf{(a)}$ The horizontal component of the velocity is constant during the whole trip since there is no acceleration in the horizontal direction so we have that the horizontal component just before the impact is the same as the initial horizontal component of the velocity i.e. $v_0=2.70\text{ m/s}$.
$\textbf{(b)}$ The vertical component is due to the gravitational acceleration. Since the initial vertical component is zero (the clam is dropped) we have that
$$v_v=gt=-9.81\text{ m/s}^2\cdot2.10\text{ s}=-20.6\text{ m/s}.$$
$\textbf{(c)}$ The answer to part $\textbf{(a)}$ would increase since the horizontal impact velocity will always be equal to the velocity of the crow.
The answer to part $\textbf{(b)}$ would remain the same since the vertical component of the velocity does not depend on the initial horizontal component of the velocity.
