Golf Ball Released from Tall Building: Velocity & Impact

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SUMMARY

A golf ball released from rest at the top of a tall building accelerates downward due to gravity at 9.8 m/s². After 3.99 seconds, the velocity of the ball can be calculated using the formula v = g * t, resulting in a velocity of approximately 39.6 m/s. Just before impact with the ground, the speed of the ball will also be 39.6 m/s, assuming no air resistance. Additionally, if a hot air balloon is descending at a constant speed of 4.9 m/s, this scenario can be analyzed in conjunction with the ball's motion.

PREREQUISITES
  • Understanding of basic physics concepts, particularly kinematics
  • Familiarity with the equations of motion under constant acceleration
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to perform unit conversions and calculations
NEXT STEPS
  • Study the equations of motion for uniformly accelerated objects
  • Learn how to calculate impact velocity in free fall scenarios
  • Explore the effects of air resistance on falling objects
  • Investigate the principles of relative motion in different reference frames
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the dynamics of free-falling objects.

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A golf ball is released from rest from the top
of a very tall building. Choose a coordinate
system whose origin is at the starting point
of the ball, and whose y-axis points vertically
upward.
The acceleration of gravity is 9.8 m/s
2
.
Neglecting air resistance, calculate the ve-
locity of the ball after 3.99 s. Answer in units
of m/s. 029 (part 2 of 4) 10.0 points
What is its speed just before impact with the
ground? Answer in units of m/s.
030 (part 3 of 4) 10.0 points
Now assume the hot air balloon is traveling
vertically downward at a constant speed of
4.9 m/s.
 
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Please submit homework in the proper format. See:
https://www.physicsforums.com/showpost.php?p=1042781&postcount=2"

Without some attempt at as solution you won't get any help here.
 
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