A ball is thrown at 30 m/s, at 37° above the horizontal…

Step 1. Split. $u_x = 24\ \text{m/s}$ and $u_y = 18\ \text{m/s}$.

Step 2. At 3.0 s. $v_x = 24\ \text{m/s}$ and $v_y = 18\ \text{m/s} - 9.8\ \text{m/s}^2 \times 3.0\ \text{s} = -11.4\ \text{m/s}$.

Step 3. Speed and direction. $\sqrt{(24\ \text{m/s})^2 + (11.4\ \text{m/s})^2} = 26.6\ \text{m/s}$, at $\tan^{-1}\dfrac{11.4\ \text{m/s}}{24\ \text{m/s}} = 25.4^\circ$ below the horizontal.