Radius of curvature of trajectory

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SUMMARY

The radius of curvature of a golf ball's trajectory, when hit with an initial velocity of 165 mph at an angle of 12 degrees above the horizon, is calculated to be 1859 feet. The discussion confirms that this radius remains unchanged when the ball descends at the same velocity but at an angle of 12 degrees below the horizon. The equation used to derive this conclusion is an = v²/ρ, where 'an' represents the centripetal acceleration, 'v' is the velocity, and 'ρ' is the radius of curvature.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with the equation an = v²/ρ
  • Knowledge of initial velocity and angle of projection
  • Basic concepts of curvature in trajectories
NEXT STEPS
  • Explore the effects of different launch angles on radius of curvature
  • Learn about the impact of air resistance on projectile motion
  • Investigate the relationship between velocity and trajectory shape
  • Study advanced projectile motion equations and their applications
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as golf enthusiasts interested in the physics behind ball trajectories.

Dusty912
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Homework Statement


a golfer hits a golf ball so that it has an initial velocity of 165mph and 12 degrees above the horizon. I already know the radius of curvature when the ball is hit. but my question is, will the radius be that same when it hits the ground?

Homework Equations


an=v2/ρ

The Attempt at a Solution



the radius of curvature when the ball is hit is 1859ft, since the velocity will be the same when it hits the ground except for it being 12 degrees BELOW the horizon, the radius should be the same right?
 
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Right.
 
thanks!
 

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