The farthest the fielder can be from where the ball will land is approximately 90.8 meters.
Determine the initial vertical velocity (v_y).
We can use the maximum height (h) and the acceleration due to gravity (g) to find the initial vertical velocity (v_y) using the following equation:
v_y^2 = 2gh
v_y = √(2 * 9.8 * 14.8)
v_y ≈ 18.3 m/s
Determine the initial horizontal velocity (v_x).
The initial velocity is 45 degrees above the horizontal, and its magnitude is unknown. We can use trigonometry to find the horizontal component (v_x):
v_x = v * cos(θ)
v_x = 45 * cos(45°)
v_x ≈ 31.8 m/s
Determine the time the ball spends in the air (t).
The time the ball spends in the air is twice the time it takes to reach its maximum height. We can use the following equation to find the time to reach the maximum height (t_max):
t_max = v_y / g
t_max = 18.3 / 9.8
t_max ≈ 1.87 s
Therefore, the time the ball spends in the air (t) is:
t = 2 * t_max
t = 2 * 1.87
t ≈ 3.74 s
Determine the horizontal distance the ball travels (x).
The horizontal distance the ball travels is equal to the product of its horizontal velocity and the total time it spends in the air:
x = v_x * t
x = 31.8 * 3.74
x ≈ 119.2 m
Determine the maximum distance the fielder can be from the landing point (d).
The fielder starts running from the moment the ball is hit and covers a distance equal to his speed multiplied by the time the ball spends in the air:
distance_fielder = speed_fielder * t
distance_fielder = 7.60 * 3.74
distance_fielder ≈ 28.42 m
Therefore, the maximum distance the fielder can be from the landing point is:
d = x - distance_fielder
d = 119.2 - 28.42
d ≈ 90.8 m