Acceleration due to gravity question

[catfish]

Hi guys

To me I sound stupid, but I'm probably just overthinking the question. I know acceleration due to gravity is 9.8m/s/s.
But if the distance is less than a metre or takes less than a second, wouldn't the acceleration be less than 9.8? If so, how is that worked out?
If not, then is it correct to say that if something falls for a distance of 0.7m and comes to a stop, the acceleration is still 9.8m/s/s?

 

[A.T.]

But if the distance is less than a metre or takes less than a second, wouldn't the acceleration be less than 9.8?

No

If not, then is it correct to say that if something falls for a distance of 0.7m and comes to a stop, the acceleration is still 9.8m/s/s?

Yes

 

[catfish]

Ok thanks. Next question. If the object bounces back up, is the acceleration still the same? And how do I go about working out the time it takes after impact to bounce to a height? The only details I've got is what I used to calculate velocity and momentum at impact.

 

[voko]

Acceleration is not defined in terms of distance and time. It is defined in terms of velocity and time. To determine average acceleration, you take initial velocity and final velocity, and divide it by time.

You can relate acceleration and distance, too, but then you need more details. For example, if you can assume that your acceleration is constant, as in the case of acceleration due to gravity close to the Earth, then you can use the SUVAT equations.

 

[/AH]

If the object bounces back up, is the acceleration still the same?

Yes, the (gravitational) acceleration is still the same (but the object would see it as a deceleration)

And how do I go about working out the time it takes after impact to bounce to a height? The only details I've got is what I used to calculate velocity and momentum at impact.

By setting up the energy balance of the system consisting of kinetic and potential energy, assuming a fully elastic bounce and no friction.
$$E = \frac{1}{2} m v^2+m g h$$
And using the height at impact (h=0) to calculate the kinetic energy of the system.
At height = h you can calculate the velocity and using $v_h = v_0 - gt$ you can calculate the time.

 

[puncheex]

Hi guys

To me I sound stupid, but I'm probably just overthinking the question. I know acceleration due to gravity is 9.8m/s/s.
But if the distance is less than a metre or takes less than a second, wouldn't the acceleration be less than 9.8? If so, how is that worked out?
If not, then is it correct to say that if something falls for a distance of 0.7m and comes to a stop, the acceleration is still 9.8m/s/s?

The acceleration due to gravity is about 9.8 m/s/s at all times and everywhere within a shell of space surrounding the sphere defined as "sea level" on Earth. Below that sphere it declines to zero (linearly? - not sure) down to the center of the Earth; above that it also declines, but exponentially, reaching zero at infinite distance.

 

[HallsofIvy]

The acceleration due to gravity is about 9.8 m/s/s at all times and everywhere within a shell of space surrounding the sphere defined as "sea level" on Earth. Below that sphere it declines to zero (linearly? - not sure)

Yes, linearly.

down to the center of the Earth; above that it also declines, but exponentially, reaching zero at infinite distance.

Not 'exponentially', proportional to the inverse square of distance.

 

[jbriggs444]

Yes, linearly.

In an ideal model where the density of the Earth is uniform, the decline would be linear. In fact, the density of the Earth is not uniform. Wikipedia has a nice graph about 2/5 of the way down the following page:

http://en.wikipedia.org/wiki/Gravity_of_Earth

 

[puncheex]

Check. Thanks for the additional information, both of you.

 

[catfish]

Ohhhh I think I know why I was confused. I was thinking about changes in velocity. So the velocity will change even though acceleration is still the same, right?