To solve this problem, we can use the physics of projectile motion. We want to find the angle at which the ball must be thrown so that it clears a 5-meter high wall, given an initial velocity of 20 m/s.

Key Equations:

In projectile motion, the vertical displacement $y$ is given by the equation:

$y = v_0 \sin(\theta) t - \frac{1}{2} g t^2$

Where:

• $y$ is the height (5 meters in this case),
• $v_0$ is the initial speed (20 m/s),
• $\theta$ is the angle of projection,
• $t$ is the time in the air at the maximum height,
• $g$ is the acceleration due to gravity (9.81 m/s²).

To solve for the angle, we first calculate the time at which the ball reaches the height of 5 meters, then use the horizontal motion to relate the time and the angle. Let's go step by step:

Step 1: Vertical Motion Equation

At the maximum height, we use:

$y = v_0 \sin(\theta) t - \frac{1}{2} g t^2$

Given $y = 5$ meters, $v_0 = 20$ m/s, and $g = 9.81$ m/s², we need to solve for $\theta$.

Step 2: Solving for the angle

Let me calculate the angle.The best angle to throw the ball so that it just clears a 5-meter high wall is approximately 86°. This angle allows the ball to reach the required height given the initial speed of 20 m/s.

Do you want more details on how the angle was determined or have any other questions?

Here are 5 related questions to expand on this:

1. What would the flight time of the ball be at this angle?
2. How far horizontally would the ball travel if thrown at this angle?
3. How would the angle change if the wall were taller or shorter?
4. What is the maximum height the ball would reach at this angle?
5. How does the mass of the ball affect its trajectory in this case?

Tip: Always ensure that both vertical and horizontal motion are considered separately in projectile motion problems!