We need to calculate the horizontal and vertical components of the launch velocity $v$, given that:

• $v = 708 \, \text{m/s}$ (magnitude of the velocity),
• $\theta = 36^\circ$ (launch angle).

The formulas for the components are:

$v_x = v \cdot \cos(\theta) \quad \text{and} \quad v_y = v \cdot \sin(\theta)$

Step 1: Calculate $v_x$ (horizontal component)

$v_x = 708 \cdot \cos(36^\circ)$

Using $\cos(36^\circ) \approx 0.809$:

$v_x = 708 \cdot 0.809 \approx 572.77 \, \text{m/s}$

Step 2: Calculate $v_y$ (vertical component)

$v_y = 708 \cdot \sin(36^\circ)$

Using $\sin(36^\circ) \approx 0.587$:

$v_y = 708 \cdot 0.587 \approx 415.56 \, \text{m/s}$

Final Results:

• $v_x = 572.77 \, \text{m/s}$ (horizontal component),
• $v_y = 415.56 \, \text{m/s}$ (vertical component).

Let me know if you'd like further clarification or additional calculations!

Related Questions:

1. How does gravity affect the projectile motion's vertical component?
2. How do we determine the total flight time for this projectile?
3. What is the projectile's range given its velocity and flight time?
4. What is the time taken to reach the maximum height?
5. How can we calculate the maximum height achieved by the projectile?

Tip:

In projectile motion, the horizontal component of velocity remains constant, while the vertical component changes due to gravitational acceleration.