To find the horizontal ($v_x$) and vertical ($v_y$) launch components of the velocity given the magnitude of the initial velocity $v$ and the launch angle $\theta$, we can use the following trigonometric relationships:

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

Where:

• $v = 500 \, \text{m/s}$ (initial velocity),
• $\theta = 30^\circ$ (launch angle).

Step 1: Calculate $v_x$

Using $v_x = v \cdot \cos(\theta)$:

$v_x = 500 \cdot \cos(30^\circ)$

From trigonometric values, $\cos(30^\circ) = 0.866$:

$v_x = 500 \cdot 0.866$ $v_x = 433.00 \, \text{m/s}$

Step 2: Calculate $v_y$

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

$v_y = 500 \cdot \sin(30^\circ)$

From trigonometric values, $\sin(30^\circ) = 0.5$:

$v_y = 500 \cdot 0.5$ $v_y = 250.00 \, \text{m/s}$

Final Answer:

• $v_x = 433.00 \, \text{m/s}$
• $v_y = 250.00 \, \text{m/s}$

Let me know if you need further details or explanations!

Related Questions:

1. How can you calculate the time of flight of a projectile?
2. What is the maximum height reached by the projectile given $v_y$ and gravity?
3. How do you determine the horizontal range of a projectile?
4. What happens to $v_x$ and $v_y$ when the launch angle changes?
5. How does air resistance affect the motion of projectiles?

Tip:

In projectile motion, the horizontal velocity $v_x$ remains constant (ignoring air resistance), while the vertical velocity $v_y$ changes due to the effect of gravity.