To determine how far the baseball drops vertically by the time it crosses the plate, we need to calculate the vertical displacement during its horizontal travel of 18.0 meters.
Since the initial velocity is directed horizontally, there is no initial vertical velocity. We can use the equation of motion for vertical displacement under constant acceleration:
Δy = (1/2) * g * t^2
where: Δy is the vertical displacement g is the acceleration due to gravity (approximately 9.8 m/s^2) t is the time taken to cross the plate
To find the time taken to cross the plate, we can use the horizontal distance and the initial horizontal velocity:
d = v * t
where: d is the horizontal distance (18.0 m) v is the initial horizontal velocity (41.0 m/s) t is the time taken to cross the plate
Rearranging the equation to solve for time:
t = d / v
Substituting the given values:
t = 18.0 m / 41.0 m/s ≈ 0.439 seconds
Now we can calculate the vertical displacement:
Δy = (1/2) * g * t^2 = (1/2) * 9.8 m/s^2 * (0.439 s)^2 ≈ 0.942 meters
Therefore, the baseball drops approximately 0.942 meters vertically by the time it crosses the plate 18.0 meters away.