The question describes a projectile motion problem. When a projectile is launched with an initial velocity $v_0$ at an angle $\theta_0$ to the horizontal, its motion can be decomposed into horizontal and vertical components:

• The initial horizontal velocity is:
  $v_{0x} = v_0 \cos \theta_0$
• The initial vertical velocity is:
  $v_{0y} = v_0 \sin \theta_0$

At the maximum height of the projectile's motion:

• The vertical velocity becomes zero because gravity causes the upward motion to stop before the projectile starts descending.
• However, the horizontal velocity $v_{0x}$ remains constant because there is no horizontal acceleration (assuming air resistance is negligible).

Thus, the speed of the projectile at the maximum height is equal to its horizontal velocity: $v_{\text{max height}} = v_0 \cos \theta_0$

Final Answer:

The speed of the projectile at its maximum height is: $v_{\text{max height}} = v_0 \cos \theta_0$

Would you like a detailed explanation of the derivation or more examples of projectile motion?

Here are 5 questions for further exploration:

1. How do we calculate the total time of flight for a projectile?
2. What is the maximum height achieved by the projectile?
3. How does air resistance affect the projectile's motion and speed at the maximum height?
4. How can we determine the horizontal range of the projectile?
5. What are the effects of varying the launch angle on the trajectory?

Tip: Remember that the horizontal and vertical motions in projectile problems are independent. Always treat them separately!