Clarification on the (+,-) of gravitational value along it's Y axis

[Truefire]

Could someone please clarify, i thought i understood but some reading has brought some confusion. As far as the gravitational value of free fall equations are concerned, is 9.8m/s or the 32ft/s (+,-) when the object is falling. The formula y=Vyo-1/2gt^2

chris

 

[tvavanasd]

Could someone please clarify, i thought i understood but some reading has brought some confusion. As far as the gravitational value of free fall equations are concerned, is 9.8m/s or the 32ft/s (+,-) when the object is falling. The formula y=Vyo-1/2gt^2

chris

The equation that you have shown is incorrect. It should be:

y = Vyo * t - 0.5 * g * t^2

The first "t" term was missing, and I added spaces and changed the "1/2" to "0.5" for clarity.

This equation is specific to gravity, not general constant acceleration. It basically assumes that positive position, velocity and acceleration are all upward, but that in this case the acceleration will be downward due to gravity.

If (for this equation) initial velocity is zero, you will find that the final position will always be a negative number since the object's final position is below it's initial position (it has gone down against the convention of up being positive).

If initial velocity is negative (directed downward), the final position will obviously also be negative.

If initial velocity is positive (directed upward), the final position will be dependent on time.

Not to create confusion, but in general for any motion calculation, you must pick a convention and stick with it. I you decide for example that positive is up for position, then the derivatives of position (velocity and acceleration) are also positive up. If we assume that positive is down, the equation would have been slightly different as would the signs of the inputs and outputs.

The output from an equation must always be measured against the conventions selected for the equation.

Hope that helps.

Mike

 

[mathman]

you should also add in y₀, the initial height.

 

[tvavanasd]

... As far as the gravitational value of free fall equations are concerned, is 9.8m/s or the 32ft/s (+,-) when the object is falling. The formula y=Vyo-1/2gt^2

chris

I just re-read the initial post.
The units of acceleration that you list are incorrect. In the cases above, they should be either m/s^2 or ft/s^2. This is important, I'm not simply being picky.

Just to be clear, based on the equations discussed, only the magnitude of gravity should be used in place of "g" in the equation. Don't re-apply the sign to "g" in calculations.

 

[pmacias]

The gravitational value, or acceleration due to gravity, can be represented by the symbol "g". This value is constant and is approximately 9.8 m/s^2 or 32 ft/s^2 on Earth. The (+,-) symbol refers to the direction of the acceleration, with a positive value indicating acceleration towards the ground and a negative value indicating acceleration away from the ground. This depends on the frame of reference, with the ground being considered as the reference point.

In the free fall equation, y=Vyo-1/2gt^2, the term "g" represents the acceleration due to gravity. This value remains constant and does not change as the object falls. The term "Vyo" represents the initial velocity of the object, which can be positive or negative depending on the direction of the initial motion. The negative sign in front of the 1/2gt^2 term indicates that the acceleration is acting in the opposite direction of the initial velocity.

Therefore, the (+,-) symbol in the equation represents the direction of the acceleration, not the value of the acceleration itself. In summary, the gravitational value of 9.8 m/s^2 or 32 ft/s^2 is always present and acts in the direction of the ground, while the initial velocity and direction of motion determine the sign in the equation. I hope this clarifies any confusion.