Answer
Horizontal Component: $v_x \approx 46.0$ ft/s
Vertical Component: $v_y \approx 38.6$ ft/s
Work Step by Step
We can visualize the velocity as a vector $\textbf{v}$ drawn from the origin (the launch point) in the first quadrant, with the x-axis parallel to the ground. We can then see $\textbf{v}$ as the hypotenuse of a right triangle with $\theta = 40^{\circ}$ adjacent to a horizontal leg $v_x$ along the x-axis (which is the x-component) and opposite a vertical leg $v_y$ parallel to the y-axis (which is the y-component). Since $|\textbf{v}| = 60$ ft/s is the magnitude of $\textbf{v}$ and thus the length of the hypotenuse, we can use trigonometry to find the components:
$v_x = |\textbf{v}| \cos \theta = 60$ ft/s $\cos 40^{\circ} \approx 46.0$ ft/s
$v_y = |\textbf{v}| \sin \theta = 60$ ft/s $\sin 40^{\circ} \approx 38.6$ ft/s
