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Acceleration due to gravity- HELP!

  • Thread starter p.ella
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so I have this assignment where we have to get the dt and vt graphs of a bouncing basketball using a motion detector. the detector is facing down above the initial height of the ball, and is always above the ball facing down. so on my dt graph, the highest point is the ground. k so I have to find the acceleration due to gravity by finding the slope of the vt graph, here's where im confused. since the positive direction in the vt graph is bouncing down to the ground, and the negative direction is coming back up to the motion detector, is it -9.8 or +9.8. Please Please please explain how, im so confused. I thought it's always -9.8, but yea i dnt understand. im in gr.11 btw. THANK YOU SO MUCH FOR ANY HELP! PLEASE REPLY ASAP! (:
 
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gneill

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If the acceleration vector points in the same direction as your coordinate y-axis, then it must be a positive value: +9.81 m/s2.
 
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@gneil- is it positive for when it's going up and down?
 

gneill

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@gneil- is it positive for when it's going up and down?
The gravitational acceleration constant is always positive. The direction of the acceleration due to gravity is always towards the ground. In your description, the ground is "up", that is, has a higher distance value than the origin (which is presumably closer to the motion detector). So for example if your x-axis is chosen to lie along the direction of motion of the ball, then the acceleration vector would be [itex] +g\hat{i} [/itex].
 
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Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

when a ball is falling down, and down is a positive direction, is the acceleration due to gravity +9.8 or -9.8

when it is bouncing back up and up is a negative direction, is the acceleration due to gravity positive or negative

Please explain. THANKKSSS (:
 
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The gravitational acceleration constant is always positive. The direction of the acceleration due to gravity is always towards the ground. In your description, the ground is "up", that is, has a higher distance value than the origin (which is presumably closer to the motion detector). So for example if your x-axis is chosen to lie along the direction of motion of the ball, then the acceleration vector would be [itex] +g\hat{i} [/itex].
so the acceleration sue to g is positive? sorry i really need you to dumb it down for me :$ and yea, the y int in the dt graph is closer to the origin
 

DaveC426913

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Re: Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

when a ball is falling down, and down is a positive direction, is the acceleration due to gravity +9.8 or -9.8

when it is bouncing back up and up is a negative direction, is the acceleration due to gravity positive or negative

Please explain. THANKKSSS (:
There is no absolute; it is simply a convention. You could do all your work with g as a -ive or as a +ive.

The critical element is that it is consistent with your other data, for example (but not limited to) initial velocity or displacement.
 

gneill

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So, in your coordinate system the ground lies in the +y direction. Gravity is directed groundward. So gravitational acceleration has a positive value in your coordinate system. An object released from rest at the origin would accelerate towards increasing values of y.

It's up to you to write equations that are consistent with your choice of coordinate axes!
 
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Re: Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

There is no absolute; it is simply a convention. You could do all your work with g as a -ive or as a +ive.

The critical element is that it is consistent with your other data, for example (but not limited to) initial velocity or displacement.
im sorry, could you dumb it down for me :$
 

DaveC426913

Gold Member
18,193
1,802
Re: Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

im sorry, could you dumb it down for me :$
Give us the problem as written.
 
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So, in your coordinate system the ground lies in the +y direction. Gravity is directed groundward. So gravitational acceleration has a positive value in your coordinate system. An object released from rest at the origin would accelerate towards increasing values of y.

It's up to you to write equations that are consistent with your choice of coordinate axes!
ohhhh that makes so much sense! thank you sooo much for all your help, I really appreciate your time! (:
 
41
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Re: Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

Give us the problem as written.
so I have this assignment where we have to get the dt and vt graphs of a bouncing basketball using a motion detector. the detector is facing down above the initial height of the ball, and is always above the ball facing down. so on my dt graph, the highest point is the ground. k so I have to find the acceleration due to gravity by finding the slope of the vt graph, here's where im confused. since the positive direction in the vt graph is bouncing down to the ground, and the negative direction is coming back up to the motion detector, is it -9.8 or +9.8. Please Please please explain how, im so confused. I thought it's always -9.8, but yea i dnt understand. im in gr.11 btw.
 
140
0
Re: Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

so I have this assignment where we have to get the dt and vt graphs of a bouncing basketball using a motion detector. the detector is facing down above the initial height of the ball, and is always above the ball facing down. so on my dt graph, the highest point is the ground. k so I have to find the acceleration due to gravity by finding the slope of the vt graph, here's where im confused. since the positive direction in the vt graph is bouncing down to the ground, and the negative direction is coming back up to the motion detector, is it -9.8 or +9.8. Please Please please explain how, im so confused. I thought it's always -9.8, but yea i dnt understand. im in gr.11 btw.
I'd say +9.81.. Since if you multiply by square of time (t^2) you will get a positive distance which you said is already defined as down.
 
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Re: Acceleration due to gravity- HELP QUICK PLEASE!!!!!!!!!!!

I'd say +9.81.. Since if you multiply by square of time (t^2) you will get a positive distance which you said is already defined as down.
cool (: thanks for your help and time, I really appreciate it!
 
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Glad to help ;) By the way the positive distance is based on assumption that the acceleration of the object starts from rest.. But i guess the main point is that the acceleration contributes to movement in the defined positive direction and is therefor also positive..
 
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Glad to help ;) By the way the positive distance is based on assumption that the acceleration of the object starts from rest.. But i guess the main point is that the acceleration contributes to movement in the defined positive direction and is therefor also positive..
yea pretty much cause the question said to determine the acceleration due to gravity by using the slope of the vt graph
 
Typical problem where a ball is thrown up (away from x axis)...I am confused because gravity is + in parts of the calculations and - in other parts. Ie: According to this book, when the ball is moving upwards, g =9.8, but when it stops moving up and the ball is coming back down g=-9.8...any ideas?
 

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