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

In summary, the problem asks for the distance traveled during the 3rd second of fall for an object released from a height of 50m. The correct formula to use is d = Vi*T + 1/2at^2, but the question is asking for the distance traveled between the 2nd and 3rd second of fall. The answer provided is 24.5m, which can be calculated using the given formula.
  • #1
wheeljeeper
2
0

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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  • #2
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.
 
  • #3
Got it.
Yep, wording was the only confusing part. Thanks a lot!
 

1. How do you calculate the distance traveled in the third second of free falling motion?

To calculate the distance traveled in the third second of free falling motion, you can use the equation d = 1/2 x g x t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds. Plug in 3 for t and solve for d to find the distance traveled in the third second.

2. What is the average velocity during the third second of free falling motion?

The average velocity during the third second of free falling motion is equal to the instantaneous velocity at the end of the second, which can be calculated using the equation v = g x t. Plug in 3 for t and solve for v to find the average velocity during the third second.

3. How does air resistance affect free falling motion?

Air resistance, also known as drag, can affect free falling motion by slowing down the object's acceleration due to gravity. This is because air resistance creates an opposing force that acts in the opposite direction of the object's motion. The larger the surface area of the object, the greater the air resistance and the slower the object will fall.

4. Can an object reach terminal velocity during free falling motion?

Yes, an object can reach terminal velocity during free falling motion. Terminal velocity is the maximum speed an object can reach while falling due to the balance between the force of gravity and the force of air resistance. As an object falls and its speed increases, the force of air resistance also increases until it is equal to the force of gravity, resulting in a constant velocity known as terminal velocity.

5. How can free falling motion be used in real-life situations?

Free falling motion is used in real-life situations such as skydiving, bungee jumping, and parachuting. It is also used in physics experiments and calculations, as well as in designing and testing objects that need to withstand free falling conditions, such as airplane emergency systems and space shuttle re-entry systems.

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