A major-league pitcher can throw a baseball in excess of 40.5 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 a certain extent, the pitcher can control how far the ball drops, or even briefly rises, by varying the top spin that he gives to the ball. They expect you to ignore aerodynamic effects, however.
The time it takes the ball to reach the catcher is
T = 17.7/40.5 = 0.437 s
Next, compute how far the ball will drop due to gravity during that time. Remember that the initial horizontal velocity component is zero.
Use the formula
change in y = (g/2) T^2
To find out how much the ball will drop by the time it reaches the catcher, we need to calculate the vertical distance it travels due to gravity during that time.
The key here is that the ball is thrown horizontally, so we only need to consider the vertical motion caused by gravity. The horizontal motion doesn't affect the vertical displacement.
Let's break down the problem into smaller steps:
Step 1: Find the time it takes for the ball to travel 17.7 m horizontally.
We know the horizontal speed of the ball is 40.5 m/s, and the distance it needs to travel horizontally is 17.7 m. We can use the formula: distance = speed × time.
So, time = distance / speed.
Plugging in the values, we get time = 17.7 m / 40.5 m/s.
Step 2: Find the vertical displacement due to gravity during this time.
Since there is no initial vertical velocity, we can use the formula for displacement due to gravity: displacement = 1/2 × acceleration due to gravity × time².
The acceleration due to gravity is approximately 9.8 m/s².
Substituting the values, we get displacement = 1/2 × 9.8 m/s² × (time from Step 1)².
Step 3: Calculate the vertical drop.
The vertical drop is the absolute value of the vertical displacement calculated in Step 2, since it's downward. So, the vertical drop = absolute value of displacement.
By following these steps, you can find the amount the ball will drop by the time it reaches the catcher who is 17.7 m away from the point of release.