To find the vertical component of the launch velocity, we can use the following formula:

$v_y = v \cdot \sin(\theta)$

Where:

• $v_y$ is the vertical component of the velocity,
• $v$ is the initial velocity (65 m/s),
• $\theta$ is the angle of launch (12°),
• $\sin(\theta)$ is the sine of the angle.

Now calculate it:

$v_y = 65 \cdot \sin(12^\circ)$

Using a calculator:

$v_y = 65 \cdot 0.2079 \approx 13.514 \, \text{m/s}$

So, the vertical component of the launch velocity is approximately 13.51 m/s.

Would you like more details on this? Here are some additional questions related to this topic:

1. How would you calculate the horizontal component of the launch velocity?
2. How does changing the launch angle affect the vertical component of velocity?
3. What role does gravity play after the arrow is launched?
4. How would you calculate the time of flight for this projectile?
5. Can the vertical velocity ever become negative during the arrow's flight?

Tip: The sine function is used for vertical components, while the cosine function is used for horizontal components when resolving vectors.