Which Distance is Correct for a Free Falling Object?

In summary, the conversation discusses the correct formula for calculating the distance traveled by a free falling object. There are two formulas presented, but only one is correct due to the fact that the velocity of the object is not constant. The correct formula is d = 1/2 a*t^2, which yields a distance of 4.9 meters for the first second of free fall. The other formula, d = vt, assumes a constant velocity and is therefore incorrect.
  • #1
arunkumarg
4
0
I have a simple question may sound stupid for most of you:

For a free falling object v=9.8 m/sec. which of the distance traveled per sec is correct:

Formula 1:

d = vt
d = 9.8/1 = 9.8 m

Formula 2:
d = 1/2 a*t^2

acc of a free falling object is 9.8m/sec^2

d = 1/2 9.8*1^2
d = 4.9 m

This is a bit confusing.
 
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  • #2


arunkumarg said:
For a free falling object v=9.8 m/sec.
No, the acceleration of a free falling object is 9.8 m/s^2. The velocity is not constant.

Formula 1:

d = vt
d = 9.8/1 = 9.8 m
This is incorrect as it assumes a constant velocity.

Formula 2:
d = 1/2 a*t^2

acc of a free falling object is 9.8m/sec^2

d = 1/2 9.8*1^2
d = 4.9 m
This is correct. That's the distance fallen in the first second.
 
  • #3


Thanks for your reply.

When I say V = 9.8m/sec. its the instantaneous velocity (for the 1st Sec).

Hence, for the 1st sec the dist must be 9.8m.

I know I am missing the point somewhere.
 
  • #4


arunkumarg said:
When I say V = 9.8m/sec. its the instantaneous velocity (for the 1st Sec).
That's the instantaneous velocity at the end of the 1st second. It starts out with zero velocity at t = 0 and ends up going 9.8 m/s at t = 1 sec.
Hence, for the 1st sec the dist must be 9.8m.
That would be true if the speed were constant for the entire second. But it's not. Since it starts from rest, the average speed during the 1st second is only 4.9 m/s.
 
  • #5


Sounds good for me... Thanks pal
 

1. What is the formula for calculating distance for a free falling object?

The formula for calculating the distance traveled by a free falling object is d = 1/2 * g * 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.

2. How does air resistance affect the distance traveled by a free falling object?

Air resistance can reduce the distance traveled by a free falling object by slowing its velocity. This is because air resistance creates a force in the opposite direction of the object's motion, which decreases its acceleration and thus reduces the distance it can travel.

3. What is the difference between displacement and distance for a free falling object?

Displacement is the straight-line distance between the starting point and ending point of an object's motion, while distance is the total length of the path traveled by the object. In the case of a free falling object, displacement and distance will be the same if the object falls straight down without changing direction. However, if the object has any horizontal motion, displacement and distance will be different.

4. Does the mass of a free falling object affect the distance it travels?

The mass of a free falling object does not affect the distance it travels. This is because, according to the formula d = 1/2 * g * t^2, the mass of the object does not factor into the equation. The only factors that affect the distance traveled are the acceleration due to gravity and the time the object has been falling.

5. How does the height from which a free falling object is dropped affect the distance it travels?

The height from which a free falling object is dropped does not affect the distance it travels. This is because, according to the formula d = 1/2 * g * t^2, the initial height of the object does not factor into the equation. The only factors that affect the distance traveled are the acceleration due to gravity and the time the object has been falling.

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