1
$\begingroup$

I have a few doubts on how the acceleration of a ball works when we throw it up in the air, In my mind, I could imagine it in some different "real world" cases, So I am listing them all down, though I suspect the same reasoning would apply to all.

  1. Say I jump out of my building's terrace (don't worry, I believe I can fly :D) and then Mid-way down, I throw a tennis ball up/down, in both cases, the instant after the ball leaves my hand, How does the acceleration of the ball compare to $g$ (ie. $>$ or $<$ or $=$)
  2. Say I am jumping on a trampoline and while going up, I throw the ball up/down, the instant after the ball leaves my hand, How does the acceleration of the ball compare to $g$ (ie. $>$ or $<$ or $=$)
  3. Say I am standing still and I throw the ball up, the instant after the ball leaves my hand, How does the acceleration of the ball compare to $g$ (ie. $>$ or $<$ or $=$)

In all of these, we are assuming there is no wind/ air friction etc. and that $g$ (gravitaional acceleration) is a fixed constant $(=-9.8)$ everywhere near Earth, so like $100 m$ up or down isn't affecting $g$

So, I was doing HRK, when 2 questions came up that were essentially asking the exact same things I am asking in case 1.

My thoughts: I thought that when I throw the ball up while falling down, the acceleration must be positive for an instant, or at least like $-9.8 + c$ (where $c$ is a positive number). My reasoning was as follows: If we throw the system onto a cartesian plane and direct up to be positive, then originally, the ball (and me) have a negative velocity, but when I throw the ball up, it has a positive velocity, which means there MUST have been an acceleration/Force. This was in agreement with the fact that in the book, when drawing the graph of a rebounding ball, they had shown a discontinuity in the acceleration, for the same reason I have described above. So, for a small instant the acceleration is $> -9.8$ but then comes back to $-9.8$ (which is what I am assuming $g$ is).

For the case when I am falling down and throwing the ball up, for a similar reason I thought that the acceleration at first must be $< -9.8$ and then come back to $-9.8$

As the Solutions to HRK's MCQ's are not given, I decided to check my answers from @knzhou 's answer key, and he stated that the Answer to both questions should be that acceleration is $-9.8$.

Now, I am $100$% sure that he must've been right (IPHO gold medallist orz), But I don't seem to understand the reason why...

About 15 months ago I had learned a little newtonian mechanics (without calculus) from an exam-prep like book, and after that, I hadn't looked at it for almost a year now, but now that I have started to learn Physics again, I want to really understand intuitively most of the things I can, So, this question may be a little elementary/ stupid, But I would still really appreciate if you could answer these.

Thanks!

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ When no external forces in addition to gravity act on the ball, the acceleration will always be $=g$. The acceleration is only different when the ball is in your hand and your hand is pushing it upwards. $\endgroup$
    – QuantumEngineer
    Commented Jun 1, 2021 at 21:18
  • 1
    $\begingroup$ @Anonymousphysicist , thank you! $\endgroup$
    – Aditya_math
    Commented Jun 1, 2021 at 21:25

1 Answer 1

Reset to default
4
$\begingroup$

It seems like you are overthinking here, since in all situations you are thinking about, the instant after the ball leaves your hand, the acceleration is $g$.

No matter if you are standing, falling or jumping while throwing a ball: Depending on your strength, you can give the ball any initial velocity you want (meaning speed and direction), which of course means you have to accelerate the ball while it is in your hand. The acceleration needed depends on the desired velocity the ball shall have, and on how long you accelerate it (meaning how long you take to throw it).

As soon as you let it go, the only force acting on it is gravity, therefore the acceleration will be $g$ after it leaves your hand.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ oh so like if they asked the acceleration at the instant I am throwing the ball, then my reasoning might be correct, but as they have asked after , the only acceleration is $g$? Thanks so much, I really appreciate your answer. $\endgroup$
    – Aditya_math
    Commented Jun 1, 2021 at 21:24
  • $\begingroup$ I do not know what you mean with instant at which you are throwing a ball. When you move your arm to throw the ball, this takes a finite amount of time. During this time, you accelerate the ball so it ends up with a certain velocity. If you want to make the throw quicker, you can still reach the same velocity, but then the acceleration need to be larger. $\endgroup$
    – Koschi
    Commented Jun 1, 2021 at 21:29
  • $\begingroup$ Oh yes, sorry for that, I think what I mean was that during the time I am moving my arms, to throw the ball, at that time the acceleration is $>g$ right? Once again, Thanks soooo much, I really appreciate you help $\endgroup$
    – Aditya_math
    Commented Jun 1, 2021 at 21:34
  • $\begingroup$ No, this is still not generally correct. Think about just holding the ball still. This means there still is gravity acting on the ball, meaning the force due to gravity is $F_g=mg$, but this is counterbalanced by a normal force on the ball by your hand. So the net force is zero, $F_{\mathrm{tot}}=0$, hence the ball does not move, i.e. the acceleration of it is 0. If you now throw the ball directly up, the force asserted by your hand must be bigger than the gravit. force, but that does not mean that the accel. is necessarily bigger than $g$, since it is calculated by $ma=F_{\mathrm{tot}}$. $\endgroup$
    – Koschi
    Commented Jun 2, 2021 at 7:02
  • $\begingroup$ oh yes, allright, that makes sense, Thanks a ton $\endgroup$
    – Aditya_math
    Commented Jun 2, 2021 at 15:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.