Why isn't the acceleration at the top point of a ball’s journey zero? When I shoot a ball vertically upward, its velocity is decreasing since  there is a downward acceleration of about $9.8\,\mathrm{ms}^{-2}$. 
I have read that at the top most point, when $v = 0$, the acceleration is still $9.8\,\mathrm{ms}^{-2}$ in the downward direction where $v=0$. That is, the acceleration is still the same.
But at the highest point, the ball is stationary, so it is not even moving. How can it accelerate?
 A: You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time:

Acceleration is defined as:
$$ a = \frac{dv}{dt} $$
so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is constant. The gradient is just the gravitational acceleration $9.81$ m/s$^2$.
The point is that the gradient, and hence the acceleration, does not depend on $v$ at all. So it is the same value of $9.81$ m/s$^2$ when $v = 0$ just as it is at all other values of $v$.
A: When you shoot the ball upwardly, gravity acts on it with a force $mg$ where $m$ is the mass of the ball and $g=9.81 ms^{-2}$ the Earth's gravitational acceleration.
If the initial upward velocity was $v_0$ then the instantaneous velocity $v$ is given by:
$v=v_0-gt$, so after some time $t=\frac{v_0}{g}$ the balls's velocity becomes $v=0$.
However, we know the ball will now start falling back to Earth immediately and if we defined $v_0$ as positive then $v=v_0-gt$ then now becomes negative. The acceleration $g$ hasn't changed though because the force $mg$ acts all the time during the trajectory.
The fact that at the apex of its path velocity becomes momentarily $0$ does not mean $g$ changes: it doesn't because the Earth's gravity acts on the ball, regardless of its velocity or elapsed time.
A: At the topmost point, the velocity vector is a null vector whereas the acceleration vector has constant magnitude $-9.8\,\mathrm{m/s^2}$ and constant direction downwards i.e. towards the centre of earth.
A: I think you are subconsciously mixing up velocity with acceleration. Let me give you an example. Imagine these are the measured speeds of a particle thrown vertically into the air at different times:
time, speed
0s,   50m/s
1s,   40m/s
2s,   30m/s
3s,   20m/s
4s,   10m/s
5s,   0m/s
6s,  -10m/s
7s,  -20m/s
8s,  -30m/s
9s,  -40m/s
10s, -50m/s <-- it hits the ground

Here a minus sign simply means the particle is coming towards the earth. Now the acceleration at for example 4s is the difference between the speeds at 3s and at 5s divided by the elapsed time, which is (20m/s-0m/s)/2s=10m/s^2
Similarly, the acceleration at 5s is the difference between the speeds at 4s and at 6s divided by the elapsed time, which is (10m/s-(-10m/s))/2s=10m/s^s
This motion is called a motion with constant acceleration, as the earth exerts a constant force on the particle at all times, no matter where it is (unless the particle gets really really far, in which case the force won't be constant any more).
A: You are correct, in that the velocity is zero, so its direction doesn't mean anything, but just because the velocity is zero doesn't mean the acceleration is zero.  And it's not zero, it's -9.8m/s/s, as you acknowledge, so the direction of acceleration is meaningful.
A: The no math answer to this one is to realize that acceleration is the rate of change of velocity. At the top most point, the velocity is indeed zero. However, it is changing momentarily after that. If the acceleration was zero, the ball would have had no change in velocity and would have stayed up in the air forever. 
A: Acceleration is the rate of change of velocity. That means it is the difference of final and initial velocity divided by the time duration between these two observations. Obviously, that means that there must be two points of time within which acceleration happens. You take any two instances of time and get the instantaneous velocities at these two instances and divide that by the interval of time, you are bound to get $9.8 \frac{m}{s^2}$.
The key point here is that while velocity is instantaneous, and therefore can be zero, acceleration is a function of the duration of time, and hence cannot be zero.
A: A projectile has zero acceleration at its peak. Yes, suppose when you drop the ball upwards direction at the maximum point. The ball has zero acceleration at this point. And then gravity pulls it back.
A: This is simple, when you throw body upwards it must have an initial velocity as u
When it reach at top final velocity v becomes 0 and body is at rest means stop
We know that 
$acc= \frac{v-u}{t}$
If $v=0$
then $acc = \frac{-u}{t}$ which is not equal to zero(0)
So it is right that there is an acc having velocity zero when body is at rest position at the top upward motion.
