Acceleration due to gravity- HELP

In summary, the conversation discusses how to find the acceleration due to gravity by using a motion detector to analyze the dt and vt graphs of a bouncing basketball. There is confusion about whether the acceleration should be positive or negative, but it is determined that in this specific scenario, the acceleration due to gravity is always positive because it is consistent with the chosen coordinate system and the direction of the ball's motion.
  • #1
p.ella
41
0
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 I am 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, I am so confused. I thought it's always -9.8, but yea i dnt understand. I am in gr.11 btw. THANK YOU SO MUCH FOR ANY HELP! PLEASE REPLY ASAP! (:
 
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  • #2
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.
 
  • #3
@gneil- is it positive for when it's going up and down?
 
  • #4
p.ella said:
@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].
 
  • #5
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 (:
 
  • #6
gneill said:
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
 
  • #7


p.ella said:
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.
 
  • #8
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!
 
  • #9


DaveC426913 said:
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 :$
 
  • #10


p.ella said:
im sorry, could you dumb it down for me :$

Give us the problem as written.
 
  • #11
gneill said:
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! (:
 
  • #12


DaveC426913 said:
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 I am 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, I am so confused. I thought it's always -9.8, but yea i dnt understand. I am in gr.11 btw.
 
  • #13


p.ella said:
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 I am 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, I am so confused. I thought it's always -9.8, but yea i dnt understand. I am 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.
 
  • #14


faen said:
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!
 
  • #15
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..
 
  • #16
faen said:
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
 
  • #17
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?
 

1. What is acceleration due to gravity?

Acceleration due to gravity, denoted as g, is the acceleration an object experiences due to the gravitational force exerted by a massive body such as the Earth. It is a constant value that is approximately 9.8 meters per second squared (m/s^2) near the Earth's surface.

2. How is acceleration due to gravity calculated?

The acceleration due to gravity can be calculated using the formula g = G*M / r^2, where G is the gravitational constant, M is the mass of the massive body, and r is the distance between the object and the center of the massive body.

3. Does acceleration due to gravity vary on different planets?

Yes, acceleration due to gravity varies on different planets depending on their mass and radius. For example, the acceleration due to gravity on the surface of Mars is 3.7 m/s^2, while on the surface of Jupiter it is 24.8 m/s^2.

4. How does air resistance affect acceleration due to gravity?

Air resistance can affect the acceleration due to gravity by slowing down the rate at which an object falls. This is because air resistance is a force that opposes the motion of an object, so it counteracts the force of gravity and reduces the object's acceleration.

5. What are the practical applications of understanding acceleration due to gravity?

Understanding acceleration due to gravity is crucial in many fields, such as physics, engineering, and astronomy. It allows us to calculate the trajectories of objects in motion, design structures that can withstand gravitational forces, and explore the laws of gravity in space.

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