Solve Free Falling Motion Homework: Traveled in 3rd Sec of Fall

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SUMMARY

The discussion focuses on calculating the distance an object falls during the 3rd second of free fall from a height of 50 meters. The correct interpretation of the problem is crucial; it asks for the distance traveled between the 2nd and 3rd seconds, not the total distance after 3 seconds. The formula used is d = Vi*T + 1/2at^2, with an acceleration of 9.81 m/s². The correct answer for the distance traveled during the 3rd second is 24.5 meters, which resolves the confusion regarding the calculations presented.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = Vi*T + 1/2at^2
  • Basic knowledge of free fall motion and gravitational acceleration (9.81 m/s²)
  • Ability to differentiate between total distance and distance traveled during specific intervals
  • Familiarity with the concept of time intervals in motion problems
NEXT STEPS
  • Review kinematic equations for motion under constant acceleration
  • Practice problems involving free fall and time intervals
  • Explore the concept of instantaneous velocity and its calculation
  • Learn about the implications of air resistance on free fall motion
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Students studying physics, educators teaching kinematics, and anyone interested in understanding free fall motion and its calculations.

wheeljeeper
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Homework Statement


An object is released from a height of 50m. How far has the object traveled during the 3rd second of fall?

Homework Equations


d = Vi*T + 1/2at^2

Doesn't work though - answer provided is 24.5m.

The Attempt at a Solution


I did a simple distance at 3 seconds with 9.81 m/s^2 as acceleration, got 29.43m. Using the formula above I got 44.45m - again, not 24.5m as provided for answer.Thanks a lot!
 
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wheeljeeper said:

Homework Statement


An object is released from a height of 50m. How far has the object traveled during the 3rd second of fall?


Homework Equations


d = Vi*T + 1/2at^2

Doesn't work though - answer provided is 24.5m.


The Attempt at a Solution


I did a simple distance at 3 seconds with 9.81 m/s^2 as acceleration, got 29.43m.
you have calculated it's speed after 3 seconds, in m/s, not its distance.
Using the formula above I got 44.45m - again, not 24.5m as provided for answer.


Thanks a lot!
This formula is the correct one to use, but the problem is asking, somewhat vaguely, not for the distance traveled after 3 seconds, but rather, the distance traveled between the 2nd and 3rd second of travel.
 
Got it.
Yep, wording was the only confusing part. Thanks a lot!
 

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