To determine the distance at which the ball will hit the field again, we need to analyze the ball's motion. Assuming there is no air resistance, we can break down the ball's initial velocity into its horizontal and vertical components.
Given: Initial velocity magnitude (v0) = 12 m/s Launch angle (θ) = 45 degrees
Calculate the initial horizontal velocity (vx): vx = v0 * cos(θ) vx = 12 m/s * cos(45 degrees) vx = 12 m/s * √2 / 2 vx = 6√2 m/s
Calculate the initial vertical velocity (vy): vy = v0 * sin(θ) vy = 12 m/s * sin(45 degrees) vy = 12 m/s * √2 / 2 vy = 6√2 m/s
The time it takes for the ball to reach its maximum height can be determined using the vertical motion equation: vy = v0y - gt, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time.
- Determine the time to reach the maximum height (t_max): vy = v0y - gt_max 6√2 m/s = 6√2 m/s - 9.8 m/s^2 * t_max
Since the vertical velocity decreases to zero at the maximum height, we have: 0 = 6√2 m/s - 9.8 m/s^2 * t_max
Solving for t_max: 9.8 m/s^2 * t_max = 6√2 m/s t_max = (6√2 m/s) / (9.8 m/s^2) t_max ≈ 0.872 s
Calculate the time of flight (t_total): Since the time of ascent is equal to the time of descent, the total time of flight can be calculated as: t_total = 2 * t_max t_total ≈ 2 * 0.872 s t_total ≈ 1.744 s
Calculate the horizontal distance traveled (d): The horizontal distance can be determined using the horizontal motion equation: d = v0x * t_total, where v0x is the initial horizontal velocity.
d = 6√2 m/s * 1.744 s d ≈ 20.78 m
Therefore, the ball will hit the field again at a distance of approximately 20.78 meters from the launch point.