How Much Does Gravity Affect a Baseball's Trajectory Over 18.4 Meters?

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SUMMARY

The discussion focuses on calculating the gravitational drop of a baseball thrown horizontally at 76.4 mi/h over a distance of 18.4 meters. The correct approach involves using the equations of motion, specifically y = 0.5 * a * t^2, where a is the acceleration due to gravity (9.8 m/s²) and t is the time taken to reach home plate (0.542 seconds). The final calculation shows that the baseball drops approximately 1.44 meters due to gravity by the time it reaches home plate.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion
  • Familiarity with kinematic equations, specifically y = y_0 + v_y t + 0.5 a t^2
  • Ability to convert units, such as miles per hour to meters per second
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the principles of projectile motion in physics
  • Learn how to convert between different units of speed and distance
  • Explore the effects of air resistance on projectile trajectories
  • Investigate advanced kinematic equations and their applications in sports science
USEFUL FOR

This discussion is beneficial for physics students, baseball coaches, and sports scientists interested in understanding the effects of gravity on projectile motion in sports contexts.

creativeone
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1. While trying out for the position of pitcher on your high school baseball team, you throw a fastball at 76.4 mi/h toward home plate, which is 18.4 m away. How far does the ball drop due to effects of gravity by the time it reaches home plate? (Ignore any effects due to air resistance and assume you throw the ball horizontally.)
2. 1. x=1/2(v0+vf)t; 2. x=v0t+1/2at^2
3. I split up the values given into a horizontal and vertical table. Since neither vertical or horizontal has time, I solved horizontal for time and got .542s. I then took that and plugged into the 2nd equation listed, and got 19.8506, which was incorrect. My vertical and horizontal table looks like the following:

Vertical- a=9.8 m/s/s, h=?, t=
Horizontal- x=18.4 m, a=0, v0=33.975 m/s, vf=33.975

Any help asap would be great thank you!
 
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Your approach looks good. Let's say everything is correct for the horizontal component...
In the vertical component you have
y = y_0 + v_y t + \frac{1}{2} a t^2
You are really just interested in how much the ball is dropping so let's say y_0 = 0 .
We'll also assume this is a perfect pitch so v_y = 0 .

So we end up with y = \frac{1}{2} a t^2 = 0.5(-9.8)(0.542^2) = -1.44 m
 

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