A major-league pitcher can throw a baseball in excess of 44.6 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 17.7 m away from the point of release?
To determine how much the ball will drop by the time it reaches the catcher, we can use the equations of motion.
First, let's break down the given information:
Initial velocity of the ball (horizontal component) = 44.6 m/s
Distance to the catcher = 17.7 m
Assuming there are no external forces acting on the ball (such as air resistance), the only force acting on the ball is gravity. As gravity affects only the vertical motion of the ball, the horizontal velocity remains constant throughout its flight.
Since the initial horizontal velocity is constant, we can use the equation:
Distance = Velocity × Time
In this case, the distance is 17.7 m, and the velocity is 44.6 m/s. Therefore, we can rearrange the equation to solve for time:
Time = Distance / Velocity
Time = 17.7 m / 44.6 m/s
Time ≈ 0.397 s
Now, we need to determine how much the ball will drop vertically during this time. We can find the vertical displacement using the equation of motion:
Vertical Displacement = (Initial Velocity × Time) + (0.5 × Acceleration × Time²)
The initial vertical velocity of the ball is zero because it was thrown horizontally. The acceleration due to gravity is approximately 9.8 m/s².
Vertical Displacement = (0 × 0.397 s) + (0.5 × 9.8 m/s² × (0.397 s)²)
Vertical Displacement ≈ 0.760 m
Therefore, the ball will drop approximately 0.760 meters or 76 centimeters by the time it reaches the catcher.