To solve this problem, we calculate the total velocity $v$ of the ball just before it hits the ground by combining the horizontal velocity ($v_x$) and the vertical velocity ($v_y$).

Given:

• Horizontal velocity: $v_0 = v_x = 20 \, \text{m/s}$
• Time of flight: $\Delta t = 3 \, \text{s}$
• Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$

Step 1: Calculate the vertical velocity ($v_y$)

Using the equation for velocity under constant acceleration: $v_y = g \cdot \Delta t$ $v_y = 9.8 \, \text{m/s}^2 \cdot 3 \, \text{s} = 29.4 \, \text{m/s}$

Step 2: Combine horizontal and vertical velocities

The total speed ($v$) is the magnitude of the resultant vector: $v = \sqrt{v_x^2 + v_y^2}$ $v = \sqrt{(20)^2 + (29.4)^2}$ $v = \sqrt{400 + 864.36} = \sqrt{1264.36}$ $v \approx 35.57 \, \text{m/s}$

Final Answer:

The speed of the ball just before it hits the ground is approximately: $\boxed{35.6 \, \text{m/s}}$

Let me know if you'd like a detailed explanation of any step or have additional questions. Here are 5 related questions for further exploration:

1. How would the speed change if the initial velocity was increased to 30 m/s?
2. What is the height of the building from which the ball was thrown?
3. How does air resistance impact the trajectory of the ball in real-world conditions?
4. What would be the flight time if the building height was known and doubled?
5. How can this problem be extended to find the angle of the ball’s velocity before hitting the ground?

Tip: Remember, the horizontal velocity remains constant in projectile motion (without air resistance), while vertical velocity changes due to gravity.