Calculating Time in gravity/acceleration problem

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SUMMARY

The discussion centers on calculating the time it takes for an object to fall a distance of ten centimeters under the influence of gravity, which is approximately 9.8 m/s². The formula used for this calculation is s = vt + gt²/2, where 's' represents displacement, 'v' is the initial velocity, and 'g' is the acceleration due to gravity. For an object dropped from rest, the time can be calculated using the formula t = (2s/g)^(1/2), resulting in t = (0.2/9.8)^(1/2) seconds. The discussion emphasizes that air resistance is neglected in this calculation.

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  • Understanding of basic physics concepts, particularly kinematics
  • Familiarity with the formula for displacement in free fall
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to perform square root calculations
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  • Study the kinematic equations for uniformly accelerated motion
  • Learn about the effects of air resistance on falling objects
  • Explore numerical methods for solving physics problems using software tools
  • Investigate the relationship between distance, time, and acceleration in various contexts
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of motion under gravity, particularly in scenarios involving short distances and free fall calculations.

jquiring
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I understand that gravity acts upon objects at a fairly steady rate on earth. How could I calculate the time in the following example:

An object falls ten centimeters before hitting the ground. How much time has passed from when it fell to when it hit the ground?

I can't use a stopwatch because the distances are so small. Since gravity is a constant (9.8 m/s2), how can I calculate the time over such a short distance? I'm sure that a computer could calculate this almost instantly, but I would like to know how to perform the calculation on paper.
 
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The general form for your problem is d=vt+gt2/2, where v is the initial speed and d is the distance. Air resistance is neglected. Assuming you simply drop it, you will get t=(0.2/9.8 )1/2 sec.
 
Last edited:
mathman said:
The general form for your problem is d=vt+gt2/2, where v is the initial speed and d is the distance. Air resistance is neglected. Assuming you simply drop it, you will get t=(0.2/9.8 )1/2 sec.

Not d, it is s for displacement :).
 

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