Acceleration

The acceleration throughout the process is the [acceleration due to gravity], which is ALWAYS POINTING DOWNWARD with CONSTANT MAGNITUDE of 9.81 m/s2.

The [arrow] that you refer to is the [velocity] of the ball, NOT the acceleration. The change of direction and magnitude of the velocity is a result of the action of the acceleration (more correctly, the action of gravitation force that gives rise to the acceleration).

In the going-up process,
velocity: points upward,
acceleration: (always) points downward
RESULT: velocity arrow is shortened gradually (represents a slowing down process) as acceleration is [opposing] velocity.

In the falling down process,
velocity: points downward,
acceleration: (always) points downward,
RESULT: velocity arrow gradually increase (indicating a speeding up process) as acceleration [helps] the velocity to increase.

Notice that at the highest point, the velocity arrow is zero (i.e. no velocity, this is why you see the ball stationary at that moment).
But the acceleration is still present and (always) points downward. This helps the velocity arrow to start increasing from zero.

Just imagine, should acceleration and velocity be both zero at the highest point, the ball would remain at that point forever because there is nothing to help the velocity arrow to increase from zero. This is, in fact, an unrealistic situation that would never happen.

Yes, you are right.

When you throw the ball upwards, you have exerted a certain force to the ball in a upward direction.

By the newton's second (?) law,
Force (N) = 'mass (kg)' times 'acceleration (m/s^2)' (F=ma)

The mass of the ball is a constant, when you exert a force on the ball (Force increases), there will be a acceleration in the upward direction on the ball too (acceleration).

But after the ball is thrown up, another force will then act on it, and that is the gravity. In the last paragraph, we have assumed that the upward direction is positive. But now gravity is a force acting on the ball in a downward direction.So, the 'Force (N)' will be a negative number, while the mass is a constant. The result is that the magnitude of the acceleration will be negative too (deceleration.) So, it is correct to say that the "arrow" will disappear at one of the moments because the ball will soon stop rising and immediately fall again.

But it is wrong to say that the acceleration is cancelled by the gravity~~~
When the ball is rising in the air, the magnitude that is declining is not the acceleration, but the velocity instead. In the law F=ma, the force acted on the ball by gravity is a constant, the mass of the ball is a constant, so the acceleration will surely be a constant too.

I study physics, and I understand that it is sometimes difficult to distinguish between 'acceleration' and 'velocity'. But you must understand the concept.

When acceleration (unit: m/s^2) = 0, velocity (unit: m/s) = a constant.
When acceleration = a constant, velocity = changing uniformly (i.e. a straight line on a velocity-time graph)

When acceleration = changing uniformly, velocity = changing with a changing rate (i.e. a curve line one a velocity-time graph)
(Try to ignore this last one if you cannot understand, it is somehow very abstract~~~ do not force yourself!!!)

2008-06-07 22:33:19 補充：
The &#39 appeared in the text are all quotation marks...

有呢個moment的...but 唔係個ball係最高時...個ball係最高時個velocity係0ms^-1 but個acceleration係=gravity..
咁個moment係幾時??
就係係個ball 向上accelerating 個陣而個acceleration慢慢減少(因為有gravity)..直到0ms^-2...呢一個moment就係no magnitude...
after 呢個moment之後...個ball就開始減速but仍然向上...因為個velocity仲係向緊上...
有咩唔明可以再問