Negative sign of acceleration [closed]

Question

This is the problem from our Physics textbook :

A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

Answer is :

x > 0 (upward and downward motion); v < 0 (upward), v > 0 (downward), a > 0 throughout;

a > 0 is what is bothering me and I have done hours of searching and trying to understand.

Kindly explain all the parts of the answer. Thanks in advance.

Answer 1

The acceleration is a vector $\mathbf{g}$ throughout the motion, and $\mathbf{g}$ is always pointing downward. Since you choose positive $x$ to be vertically downward, so $\mathbf{g}$ is along positive $x$ if we draw out the Cartesian coordinate, then $\mathbf{g}$ must have positive value, $\mathbf{g}=g\,\hat{\mathbf{x}}$. If you choose vertically upward to be $x>0$, then acceleration $\mathbf{g}$ has negative sign, $\mathbf{g}=-g\,\hat{\mathbf{x}}$. It's just the matter how you choose the $x>0,y>0$ directions. Draw a diagram of $v$, $g$, force on the ball with $(x,y)$ coordinates.

Answer 2

a > 0 is what is bothering me

The only acceleration in the problem is the acceleration of gravity, correct?

If you agree with that, then I think you'll agree that the acceleration is always pointing downward towards the Earth and thus, the acceleration does not change sign for this problem.

Now, from the problem statement:

vertically downward direction to be the positive direction

So, what is the sign of the acceleration?