What is the final velocity of the golf ball before it lands?

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SUMMARY

The final velocity of a golf ball before it lands, after being struck at a speed of 17.1 m/s at an angle of 52.0˚, is determined using kinematic equations. The initial vertical velocity component is calculated as 13.47 m/s. The maximum height reached by the ball is 9.81 m, and the final vertical velocity before landing is calculated using the equation v² = u² + 2as, resulting in a speed of 11.07 m/s. The discussion highlights the importance of considering both vertical and horizontal components of velocity for accurate calculations.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of projectile motion concepts
  • Ability to decompose velocity into horizontal and vertical components
  • Familiarity with gravitational acceleration (9.81 m/s²)
NEXT STEPS
  • Study the derivation and application of kinematic equations
  • Learn about projectile motion and its equations
  • Explore the effects of air resistance on projectile motion
  • Investigate the calculation of horizontal velocity components in projectile motion
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Physics students, educators, and anyone interested in understanding projectile motion and kinematics in sports contexts.

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A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.1 m/s at an angle of 52.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

Initial velocity = 17.1sin52=17.1 x 0.788
=13.47 m/s.

max height = 9.81

v^2=u^2 + 2as

v^2=0+2*9.81*6.25

v= 11.07 m/s

My answer is wrong.
 
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fedora said:
A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.1 m/s at an angle of 52.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

Initial velocity = 17.1sin52=17.1 x 0.788
=13.47 m/s.

max height = 9.81

v^2=u^2 + 2as

v^2=0+2*9.81*6.25

v= 11.07 m/s

My answer is wrong.
looks like you calculated max height as 9.25 not 9.81.
An extra step, but you correctly got the y component of the velocity. What about the x component?
 

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