Determine the period of the motion

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SUMMARY

The discussion focuses on calculating the period of motion for a ball dropped from a height of 4.00 m, which undergoes an elastic collision with the ground. The correct period (T) is determined to be 1.81 seconds, derived from the formula T = 2√(2h/g), where h is the total distance traveled (8 m) and g is the acceleration due to gravity (9.8 m/s²). The initial miscalculation of 1.278 seconds resulted from only considering the time for the ball to fall and rise back to the original height, neglecting the total distance in one complete cycle.

PREREQUISITES
  • Understanding of kinematic equations, specifically y = 1/2 at²
  • Knowledge of gravitational acceleration (g = 9.8 m/s²)
  • Familiarity with the concept of elastic collisions
  • Basic grasp of periodic motion and its characteristics
NEXT STEPS
  • Study the derivation of the period formula for simple harmonic motion
  • Explore the effects of air resistance on the period of motion
  • Learn about energy conservation in elastic collisions
  • Investigate the relationship between height and period in free-fall scenarios
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of motion and energy conservation in elastic collisions.

aliciagu
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Out of ideas, and frustrated!

Homework Statement


A ball dropped from a height of 4.00 m makes an elastic collision with the grond. Assume that no mechanical energy is lost due to air resistance. Determine the period of the motion


Homework Equations



I know the period (T) = the time required for the object to make one cycle.

The Attempt at a Solution



I get: 4m/9.8 m/s^2=.408 s^2 s=.639 (square root of .408) .639 x 2= 1.278 s (time for it to go down 4 m and up 4 m)
My book gives 1.81 s as the answer. I can get that if I take the square root of 8/9.8 (8 being the total distance traveled in one period) and multiply it by 2. But why? That would be the time required for 2 periods, wouldn't it? I'm banging my head againstthe wall on this one!
 
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you can calculate the time for it to reach the ground from the kinematic eq:

y = 1/2 at^2. The time for this will be the way down, or 1/2 the cycle.
 
That answers it! I knew I was missing something stupid!
 

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