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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.

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closed as too localized by dmckee Nov 7 '12 at 3:30

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

We don't do particular instance of homework questions. There is a meta thread that explains how to ask basic questions in a way that we will answer. – dmckee Nov 7 '12 at 3:30
up vote 3 down vote accepted

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.

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Since top is x=0, so position remains positive through out, and why velocity sign is - when going up and + when going down, as object never enters negative territory which is beyond x=0, so entire action takes place in the plus zone. – pokrate Nov 7 '12 at 5:37

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?

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