# Motion/Acceleration: free-fall gravity vs. ball being push downward

Motion/Acceleration: free-fall gravity vs. ball being "push" downward

## Homework Statement

If by accident a ball were given an initial downward push instead of being freely released, would the resulting g be different? Explain.
(this is ignoring any air resistance)

N/A

## The Attempt at a Solution

Being freely released, gravity is the only factor being considered so the acceleration being consistently acted upon it is 9.8 m/s2. But if being pushed downward, the ball would now have even more acceleration from the extra "force", correct? so there would definitely be a different resulting g between the two.

Is my wording correct? In my first Physics class and am not too comfortable with the lingo yet. I am also a little confused on what 'g' is actually referring to. Is 'g' referring to gravity or just acceleration in-general? Because if meaning acceleration, then I would assume what I said is correct above. Any assistance in helping me correctly conceptualize this information?

Thanks guys!
nixm01

Pengwuino
Gold Member

'g' specifically means the gravitational acceleration of an object at the surface of the Earth, 9.80m/s^2. A general acceleration is simply denoted 'a'.

I believe the problem implies that the force does not act on the object constantly. All it says is that at the start, it's thrown downward. So imagine someone at the top of a skyscraper with a ball. He throws it downward. The force from the person acts just for the fraction of a second to give it an initial velocity downward. The problem seems to be asking what happens after the ball has been released. So in short, the problem is basically asking if the gravitational acceleration, g, depends on the initial velocity.

I would assume yes that it would have an affect on the resulting 'g' because it was given that extra acceleration down - I guess as a boost. Am I correct in assuming that any extra acceleration given to an object will no longer be acted upon by gravity alone, thus resulting in a higher 'g'?

That seems to be the most obvious assumption, but I am not sure if there is another "law" that states other-wise. To my surprise, the study of motion/acceleration/gravity has actually thrown a few curve balls at me with things I would have never thought about/known.

Another thought I am having is wondering if the question is needing answer describing what you stated - that although the beginning acceleration will be higher due to the push, it will eventually go back to being only acted upon by gravity by the time it hits the ground - which seams plausible. BUT, wouldn't that depend on how high up the ball is from the ground? Because if the above assumption is true and it will eventually get back to 9.8 m/s2, if I am just throwing the ball down from a standing position, by the time the ball hit the ground, the resulting g would still be higher since it had not had any time to change (again, only if I was correct in my assumption).

Thanks guys

Pengwuino
Gold Member

Another thought I am having is wondering if the question is needing answer describing what you stated - that although the beginning acceleration will be higher due to the push, it will eventually go back to being only acted upon by gravity by the time it hits the ground - which seams plausible. BUT, wouldn't that depend on how high up the ball is from the ground? Because if the above assumption is true and it will eventually get back to 9.8 m/s2, if I am just throwing the ball down from a standing position, by the time the ball hit the ground, the resulting g would still be higher since it had not had any time to change (again, only if I was correct in my assumption).

Thanks guys

No. The instant the ball leaves the persons hand, the acceleration becomes g = 9.80m/s^2.