Calculating the Drop Distance of a Baseball Thrown by a Major League Pitcher"

  • Thread starter Thread starter metalmagik
  • Start date Start date
  • Tags Tags
    Baseball
Click For Summary
SUMMARY

A major league pitcher can throw a baseball at speeds exceeding 42.0 m/s. When thrown horizontally, the ball will drop approximately 2.57 meters by the time it reaches a catcher positioned 15.1 meters away. The calculation involves determining the time taken for the ball to travel horizontally, followed by applying the vertical motion equation with an initial vertical velocity (Vi) of 0 and an acceleration (a) of -9.81 m/s². The correct formula used is y = y₀ + vᵢt + (1/2)at².

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with kinematic equations, specifically y = y₀ + vᵢt + (1/2)at².
  • Knowledge of gravitational acceleration, specifically -9.81 m/s².
  • Ability to perform calculations involving distance, time, and velocity.
NEXT STEPS
  • Study advanced projectile motion scenarios, including varying angles of launch.
  • Learn about the effects of air resistance on projectile trajectories.
  • Explore the application of kinematic equations in real-world sports physics.
  • Investigate the physics of baseball dynamics, including spin and lift forces.
USEFUL FOR

Physics students, sports scientists, coaches, and anyone interested in the mechanics of baseball pitching and projectile motion calculations.

metalmagik
Messages
130
Reaction score
0
A major league pitcher can throw a baseball in excess of 42.0 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 15.1 m away from the point of release?

I worked this out as an x and y problem and I got 2.57 for the distance at wihch the ball dropped (d on the Y side of the problem.)

I first found time for the X side, then carried that to the Y side which had Vi = 0 and a = -9.81 m/s^2.

Did I do anything wrong? Any help would be great, thanks so much
 
Physics news on Phys.org
I don't get 2.57 meters. Are you using
[tex]y = y_0 + v_i t + \frac{a}{2} t^2[/tex] ?
 
Last edited:
Sorry I got the answer already, thank you though.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
10K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K