How High Does a Golf Ball Go When Driven at 24.5 Degrees?

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SUMMARY

The maximum height attained by a golf ball driven at an angle of 24.5 degrees can be calculated using the formula H = (R/4) tan(θ), where R is the range of the projectile. Given that the range R is 310.24 m, the maximum height H can be determined as H = (310.24 m / 4) tan(24.5°). This results in a maximum height of approximately 10.57 m. The calculations utilize fundamental projectile motion equations, specifically focusing on the relationship between range, angle, and height.

PREREQUISITES
  • Understanding of basic physics concepts related to projectile motion
  • Familiarity with trigonometric functions, specifically sine and tangent
  • Knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Ability to perform calculations involving angles in degrees
NEXT STEPS
  • Study the derivation of projectile motion equations in physics
  • Learn how to apply trigonometric identities in real-world scenarios
  • Explore the effects of launch angle on projectile range and height
  • Investigate advanced projectile motion simulations using software tools
USEFUL FOR

This discussion is beneficial for physics students, golf enthusiasts interested in the mechanics of the game, and educators teaching projectile motion concepts.

richardjoe05
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A golfer drives a golf ball 310.24 m down the fairway. If the ball is launched at an angle of 24.5 o to the horizontal, what is the maximum height attained by the ball during its flight?


H=1/2gt^2
 
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The maximum height attained is given by H = \frac{u^2 \sin^2 \theta}{2 g} where u is the magnitude of the initial velocity provided and θ the angle with the horizontal.

The range of the projectile is R = \frac{u^2 \sin 2 \theta}{g} which is given to be 310.24 m.

Thus we have H = \frac{R}{4} \tan \theta
 

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