To solve this problem, we need to analyze the horizontal and vertical components of the velocity of the ball separately.

Given:

• Initial horizontal speed $v_{x0} = 31 \, \text{m/s}$
• Time $t = 22 \, \text{s}$
• Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$
• Initial vertical speed $v_{y0} = 0 \, \text{m/s}$ (since the ball is thrown horizontally)

Horizontal Velocity:

The horizontal velocity remains constant because there is no horizontal acceleration (assuming no air resistance). $v_x = v_{x0} = 31 \, \text{m/s}$

Vertical Velocity:

The vertical velocity is affected by gravity. The vertical velocity after time $t$ is given by: $v_y = v_{y0} + g \times t$ Since $v_{y0} = 0 \, \text{m/s}$, we have: $v_y = 0 + (9.8 \, \text{m/s}^2) \times 22 \, \text{s}$ $v_y = 215.6 \, \text{m/s}$

Final Answer:

• Horizontal velocity after 22 seconds: $v_x = 31 \, \text{m/s}$
• Vertical velocity after 22 seconds: $v_y = 215.6 \, \text{m/s}$

Would you like any further details or explanations? Here are 5 related questions:

1. How do you calculate the time of flight for a projectile launched horizontally?
2. What is the formula to calculate the range of a projectile?
3. How does air resistance affect the motion of a projectile?
4. Can you explain the concept of relative velocity in projectile motion?
5. What is the significance of the angle of projection in projectile motion?

Tip: In projectile motion, always break the motion into horizontal and vertical components to simplify the analysis.