Is the Acceleration of a Thrown Ball Upward or Downward Negative or Positive?

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SUMMARY

The discussion centers on the sign convention for acceleration when throwing a ball upward or downward, specifically the role of gravitational acceleration (g). Participants agree that while conventionally upward is positive and downward is negative, the choice of convention does not affect the final outcome as long as it is consistently applied. The acceleration due to gravity is always directed downward, and its magnitude is 9.81 m/s². Correct application of the equations of motion, such as s = ut + 1/2at², is crucial for accurate results.

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When we throw a ball upwarsd is its acceleration which is g negative or positive?
Please tell me the reason and prove it with equation.
What i think is it should be negative because the ball is going against gravity.

When we throw a ball downward is its acceleration which is g negative or positive?
I think it should be positive.


Can anybody correct me?
 
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By convention downwards is ussually stated to be a 'negative' direction whereas upwards is positive, does that answer your question?

If the ball is moving upwards by convention it has a positive velocity, if it is slowing down, which direction is it is accelerating?

But in the end it depends entirely on which way you want to call positive, is this really a homework question? Seems kinda silly.
 
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And for the second case where the ball is thrown downwards, if we agree to call downwards negative, in which direction is it accelerating?

Gravity cannot both act in a positive and negative direction! (assuming you keep your sign convention the same...)
 
OK. The problem is when i am solving question regarding throwing of a ball upward, if i use -g in my equation of motion the answer is right but conventially i should use +g, which will bring a wrong answer. !:confused:
 
Just use the magnitude of g = 9.81m/s^2 and correct for the sign convention afterwards.

How would it bring the wrong answer? Can you demonstrate?

The only thing it would do is change the sign of your answer. If you are worried about being clear about which direction an acceleration acts, just replace the + or - with upwards or downwards.
 
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Ok I see what you mean by it would bring the wrong answer. The first step to solving any problem is to set up a sign convention.

Let me demonstrate with an example problem:

Find the speed of a baseball thrown upwards with a velocity of 5.0m/s 5 seconds after it leaves the pitcher's hand. Assuming he throws it above a bottomless pit so it does not stop mid-flight.

Answer:

Just to demonstrate that the positive/negative directions are merely CONVENTIONS and it doesn't matter which way you call positive or negative, lets call upwards negative and downwards positive.

v_{i}=-5.0m/s
a=g=+9.81m/s^{2} (gravity acts as a downwards acceleration, hence a positive acceleration under this sign convention)

v_{f}=v_{i}+a*t
v_{f}=-5.0m/s+9.81m/s^{2}*5.0s
v_{f}=-5.0m/s+49.05m/s
v_{f}=+44.05m/s

Since we designated downwards as the positive direction, we conclude that after 5.0s the baseball is moving with a velocity of 44.1m/s downwards. Get it?
 
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A startled armadillo leaps upward rising 0.536 m in the first 0.190 s.
(a) What is its initial speed as it leaves the ground?
s=vit+1/2gt^2
1.89(wrong)--but when i used s=vit-1/2gt^2 the answer was correct that is
[3.75] m/s

Same thing happened with all other questions?
But when i read book and all internet souces they say when something is thrown upwards its a=+g
 
bilalbajwa said:
A startled armadillo leaps upward rising 0.536 m in the first 0.190 s.
(a) What is its initial speed as it leaves the ground?
s=vit+1/2gt^2
1.89(wrong)--but when i used s=vit-1/2gt^2 the answer was correct that is
[3.75] m/s

Same thing happened with all other questions?
But when i read book and all internet souces they say when something is thrown upwards its a=+g

Okay but you see why you got it wrong right? Because if you are saying that the initial velocity is positive, then you cannot also say that the gravitational acceleration is positive, since it acts down in the opposite direction it is a negative acceleration if you want to call up positive.

If you were to call your initial upward velocity negative, then the downward gravitational acceleration would be a positive acceleration.
 
bilalbajwa said:
A startled armadillo leaps upward rising 0.536 m in the first 0.190 s.
(a) What is its initial speed as it leaves the ground?
s=vit+1/2gt^2
1.89(wrong)--but when i used s=vit-1/2gt^2 the answer was correct that is
[3.75] m/s

Same thing happened with all other questions?
But when i read book and all internet souces they say when something is thrown upwards its a=+g

The gravitational force acts downwards, and thus the acceleration due to gravity will always be downwards. As circles says, it matters only that you pick a sign convention and stick to it. The correct equation that you should know is s=ut+1/2at^2, where a is the acceleration of the particle.

Now, for this question I would assume that you take u to be positive (positive initial speed) and s to be positive (positive displacement). Note that these are both measured upwards. Therefore, the acceleration which is acting downwards should be negative.

The best thing to do for all these sorts of problesm is to draw a diagram. Mark on your sign convention with arrows, then you are less likely to get confused.
 
  • #10
Sorry i am unable to understand.
Now When i use s=-vit+1/2gt^2 answer is still wrong!
 
  • #11
bilalbajwa said:
Sorry i am unable to understand.
Now When i use s=-vit+1/2gt^2 answer is still wrong!

You need to set s to be negative too :wink:
 
  • #12
OK I think i am getting what you guys are saying:
when we throw something in upward direction we can use either of these equations
-s=-ut+1/2gt^2 or
s=ut-1/2gt^2

Am i right!
 
  • #13
Yes! But of course you see that they are the exact same equation just each side is multiplied by -1 :biggrin:

The important thing is that negative and positive are just conventions, and as long as you pick one and stick to it throughout the entire problem like cristo says, you will arrive at the correct answer ;-).
 
  • #14
Thanks a lot guys. I get a very very very very very quick response here. Thanks a lot.
:smile::smile::biggrin:
 

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