Why Is Displacement Only 4.9m After 1 Second in Free Fall?

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SUMMARY

The displacement of a free-falling object after 1 second is 4.9 meters, calculated using the formula s = 1/2 * a * t², where acceleration (a) is 9.8 m/s². This formula provides the total displacement from the start, not the distance fallen in each second. For example, after 2 seconds, the total displacement is 19.6 meters. Understanding this distinction is crucial for accurate calculations in kinematics.

PREREQUISITES
  • Understanding of basic kinematic equations
  • Knowledge of acceleration due to gravity (9.8 m/s²)
  • Familiarity with the concept of displacement
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the kinematic equation s = ut + 1/2 at² in detail
  • Learn how to calculate displacement in different time intervals
  • Explore the effects of initial velocity on displacement
  • Review examples of free-fall motion and related calculations
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Students of physics, educators teaching kinematics, and anyone interested in understanding the principles of motion and displacement in free fall.

mirza21
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Hi friends am new to Physics. Am facing a problem while caculating displacment.

Acceleration of free falling object is 9.8m/s/s. Am using formula 1/2*b*h to get the displacement.According to me if object acceleration is 9.8 m/s/s then after 1 second it's displacement should be 9.8 meters and after 2 second it should be positioned at 29.4 m.
If i use the formula to calculate displacement then it is positioned at 4.9 m at end of 1st second and a the end of 2nd second it 19.6 m. Please explain why object is positon at 4.9 m by using this formula.


At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m

At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

At t = 5 s

d = (0.5) * (10 m/s2) * (3 s)2 = 44.1 m

For refrence see http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5d.html
 
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mirza21 said:
Hi friends am new to Physics. Am facing a problem while caculating displacment.

Acceleration of free falling object is 9.8m/s/s. Am using formula 1/2*b*h to get the displacement.According to me if object acceleration is 9.8 m/s/s then after 1 second it's displacement should be 9.8 meters and after 2 second it should be positioned at 29.4 m.
If i use the formula to calculate displacement then it is positioned at 4.9 m at end of 1st second and a the end of 2nd second it 19.6 m. Please explain why object is positon at 4.9 m by using this formula.


At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m

At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

At t = 5 s

d = (0.5) * (10 m/s2) * (3 s)2 = 44.1 m

For refrence see http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5d.html

The statement in red is totally wrong.It also depends on the initial velocity of the object from which it was thrown downwards.
NOW for the hint part:

It seems you are confused with the formula part.
The formula s=1/2 at^2 [u=0] does not give the displacement in that particular time interval,in fact It gives you the total displacement from the initial time interval.
To get the displacement in a particular time interval you must calculate the displacements using the formula and substituting the correct values and then calculate the difference.
 
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