Baseball Slugger's Pitch: Will it be a Home Run?

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SUMMARY

The discussion centers on a physics problem involving projectile motion, where a baseball slugger hits a ball 4 feet above the ground with an initial velocity of 125 ft/s at a slope of 3 vertical to 4 horizontal. The calculations reveal that the ball reaches the center-field fence, located 455 feet away, at approximately 4.55 seconds. While the user initially calculated the ball's height at the fence to be 243 feet, which would clear the 10-foot fence by 233 feet, the textbook answer states that the ball clears the fence by only 2 feet, indicating an error in the user's calculations.

PREREQUISITES
  • Understanding of projectile motion and kinematics
  • Familiarity with the equations of motion, particularly F=ma
  • Knowledge of gravitational acceleration values (9.81 m/s² or 32 ft/s²)
  • Ability to manipulate and solve quadratic equations
NEXT STEPS
  • Review the principles of projectile motion in physics
  • Learn how to apply the equations of motion to real-world scenarios
  • Practice solving problems involving gravitational acceleration
  • Explore the differences between using metric and imperial units in physics calculations
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding projectile motion and its applications in sports dynamics.

xzibition8612
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Homework Statement



A baseball slugger connects with a pitch 4 ft above the ground. The ball heads toward the 10-ft-high center-field fence, 455 ft away. The ball leaves the bat with a velocity of 125 ft/s and a slope of 3 vertical to 4 horizontal. Neglecting air resistance, determine whether the ball hits the fence (if it does, how high above the ground?) or whether it is a home run (if it is, by how much does it clear the fence?).

Homework Equations



F=ma (Euler form)

The Attempt at a Solution


The book gives the answer as home run, clears fence by 2 ft.

-mgj = m[ x2i+y2j+z2k) Note that i,j,k are unit vectors, and x2, y2, and z2 means double dot or acceleration.

-g = y2
0 = x2

y1 = -gt + C [C=125*(3/5)] => y1 = -gt + 75 (y1 is y dot or velocity)
y = (-gt^2)/2 + 75t + C [C=4] => y = -(gt^2)/2+75t+4

x1 = C [C=125*(4/5)] => x1 = 100
x = 100t + C [C=0 because no movement in x direction at t = 0] = > x=100t

Got these equations, then plugged in 455 ft into x=100t to get t=4.55s. So at 4.55s the ball would reach the fence. As to how high it will be, plug t=4.55 into the y equation, and my answer is 243 ft. The fence is 10 ft, so it clears it by 233 ft. But the book's answer is 2 ft. I know I screwed up somewhere, but I can't see it. Any help would be appreciated.
 
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hi xzibition8612! :smile:

(btw, pleeease don't write x1 x2 etc, write x' x'' :wink:)

did you use 9.81 for g, or 32 ?
 
wow :cry:
 

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