Determine the period of the motion

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The discussion revolves around calculating the period of motion for a ball dropped from a height of 4.00 m that undergoes an elastic collision with the ground. The initial calculations suggest a period of 1.278 seconds, but the correct answer is 1.81 seconds. The confusion arises from the need to account for the total distance traveled in one complete cycle, which is 8 m (down and up). The correct approach involves using the kinematic equation to find the time for the ball to fall and then doubling it for the full cycle. Ultimately, recognizing the total distance traveled clarifies the solution.
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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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