Why is gravity considered a negative vector in a pendulum question?
I hope you are familiar with Cartesian Coordinate System
According to this system, we take any vector directed downward and leftward to be negative, and any vector directed upward or rightward to be positive. The signs represent the direction of the vector(*).
Since gravity always acts downward, towards the center of the Earth, we take it to be negative.
However it is not mandatory to use this system. You can take upward direction to be negative. Then you have to take the opposite direction as positive(*).
It may be pointed out here that gravity vector has to be strictly represented by $-g\widehat{i}$ when Cartesian system is being used, where $\widehat{i}$ represents a vector of unit magnitude along positive y-axis(so $-\widehat{i}$ is a unit vector along negative y-axis). It is also possible to use define and use your own unit vectors to represent directions. When vectors along the same direction are equated the unit vectors are generally neglected.
$*$ Just to make things explicitly clear, vectors are not positive or negative. It's their direction which is represented by the signs. Oppositely directed vectors (belonging to same quantity) tend to cancel each other out. Hence to make our math and lives less complicated, if we take one direction as positive, we take the other direction negative so that we can just add the vectors and get the resultant vector.
As a matter of fact, we all know that gravity is downward (or more accurately towards the center of the earth).
$$\mathbf{g}=-\frac{GM_e}{R_e^2}\hat{r}$$
if we negate that vector, then it should move the ball up.
If there were no gravity (or that means you are in space), then there will no force, and the pendulum will remain in its initial position. If you give it some initial tangential velocity, it will set into circular motion.
If you have a problem with the negative sign, then it's a convention of the coordinate system to take down to be negative and up to be positive. But it's not necessary, you can choose your convension, the result will remain the same.
Your question addresses the direction of gravitational force on the pendulum, but there is a deeper idea that defines the direction of gravity. The concept of the gravitational force that mass 1 exerts on mass 2 is inherently the concept of a force vector, made explicit by a mathematical formulation that references a unit vector pointing from mass 2 to mass 1:
$\vec{F_{1,2}}=\frac{Gm_{1}m_{2}}{r^{2}}\hat{r}_{2,1}\:\:\:\:$(1)
This formula is a recipe for drawing the force vector that mass 1 exerts on mass 2. The unit vector $\hat{r}_{2,1}$ always points in the direction from mass 2 to mass 1. You should confirm for yourself that with this definition of the unit vector, we can switch the labels of mass 1 and mass 2 to find the force on the other mass, and that the force on the other mass will also be attractive.
Notice that we could also write the formula as
$\vec{F_{1,2}}=-\frac{Gm_{1}m_{2}}{r^{2}}\hat{r}_{1,2}\:\:\:\:$(2)
where the unit vector points from mass 1 to mass 2. Notice the trade-off in this definition: it preserves a typographical convention of typesetting each mass 1 reference before the counterpart mass 2 reference, but it now requires you to mentally flip the sign of vector using the minus sign. (If we didn't flip the sign, the force would be repulsive, which is the general structure for force interactions between like electrical charges).
The two vectors produced by both equations are identical, because in space they have the same length and point in the same direction.
Near the earth's surface where we assume constant gravitational acceleration, we can write the equation for downward force on the pendulum a few different ways, but they all yield the same vector:
$\vec{F_{g}}=\:\:\:mg\hat{y}\:\:\: $where$ \:\:\:(g = -9.8\: m/s^{2})\:\:\:\:$ (3)
$\vec{F_{g}}=-mg\hat{y}\:\:\: $where$ \:\:\:(g = 9.8\: m/s^{2})\:\:\:\:\:\:\:\:$ (4)
$\vec{F_{g}}=\:\:\:m\vec{g}\:\:\: $where$ \:\:\:(\vec{g} = -9.8\: m/s^{2}\:\:\hat{y})\:\:\:$ (5)
$\vec{F_{g}}=-m\vec{g}\:\:\: $where$ \:\:\:(\vec{g} = 9.8\: m/s^{2}\:\:\hat{y})\:\:\:\:\:\:$ (6)
Note that equation 6, while generating the correct force vector, can be misleading as vector $\vec{g}$ is positive. Physically this equation is still self-consistent, as from Newton's 2nd Law, $\vec{a}$ = $\vec{F} / m $. In equation 6, the vector $\vec{g}$ is NOT the acceleration, but rather the negative of the acceleration:
$\vec{a}$ = $\vec{F_{g}} / m $
$\vec{a}$ = $-m\vec{g} / m $
$\vec{a}$ = $-\vec{g} $
I point this out because it is not uncommon to see teachers use equation 6. My preference is to avoid using equation 6 due to the possibility of confusion about the direction of acceleration.