Final answer:
The softball's horizontal displacement is approximately 70.9 m, and the vertical displacement from the ground is approximately 2.7 m after considering gravity and the initial height.
Step-by-step explanation:
The question asks to find the horizontal and vertical displacements of a softball hit at a specific angle and velocity, considering gravity and initial height. To solve for the displacement of a projectile under the influence of gravity, we'll use the equations of motion.
First, we determine the horizontal and vertical components of the initial velocity. The horizontal velocity (vx) is the cosine of the angle times the total velocity, and the vertical velocity (vy) is the sine of the angle times the total velocity.
- vx = 28.6 m/s * cos(60.5°) = 28.6 m/s * 0.500 = 14.3 m/s (approximately)
- vy = 28.6 m/s * sin(60.5°) = 28.6 m/s * 0.868 = 24.8 m/s (approximately)
The horizontal displacement (x) is simply the horizontal velocity multiplied by the time:
x = vx * t = 14.3 m/s * 4.96 s = 70.9 m (approximately)
To find the vertical displacement (y), we use the vertical motion equation:
y = vy * t - (1/2) * g * t^2
Substituting the given values:
y = 24.8 m/s * 4.96 s - (1/2) * 9.81 m/s2 * (4.96 s)^2
y = 123.0 m - 120.8 m = 2.2 m
However, since the ball was hit from an initial height of 0.500 m, the total vertical displacement from the ground is:
Total vertical displacement = initial height + vertical displacement = 0.500 m + 2.2 m = 2.7 m