To find the vertical component of the projectile's velocity at the maximum height of its trajectory, we can use principles from projectile motion.

Given Values:

• Initial velocity ($v_0$) = 45.0 m/s
• Launch angle ($\theta$) = 65.0°
• Gravitational acceleration ($g$) = 9.8 m/s²
• Initial height ($h_0$) = 0 m (the projectile starts from ground level)

Step 1: Find the vertical component of the initial velocity

The initial velocity is split into two components: horizontal and vertical. The vertical component ($v_{0y}$) is given by:

$v_{0y} = v_0 \cdot \sin(\theta)$

Substitute the values:

$v_{0y} = 45.0 \cdot \sin(65.0^\circ)$

Using $\sin(65.0^\circ) \approx 0.9063$:

$v_{0y} \approx 45.0 \cdot 0.9063 = 40.8 \, \text{m/s}$

Step 2: Find the vertical velocity at the maximum height

At the maximum height of the projectile's trajectory, the vertical component of the velocity becomes zero. This is because at the peak, the projectile momentarily stops moving upward before it starts descending. Hence:

$v_y = 0 \, \text{m/s}$

Thus, the vertical component of the velocity at the maximum height is 0 m/s.

Summary:

The vertical component of the velocity at the maximum height of the trajectory is 0 m/s.

Would you like further details or any clarification on the calculations?

Here are 5 related questions:

1. How do you calculate the time of flight for a projectile?
2. How do you find the maximum height of a projectile?
3. What is the horizontal component of a projectile's velocity?
4. How do you determine the range of a projectile?
5. What would change if the launch angle were different?

Tip: Remember, at the maximum height of a projectile, the vertical velocity is always zero, but the horizontal velocity remains constant throughout the flight (in the absence of air resistance).