when I shooting a ball vertically upward, its velocity is decreasing since there is a bottom acceleration of about \$9.8,mathrmms^-2\$.

I have read the at the top most point, as soon as \$v = 0\$, the acceleration is quiet \$9.8,mathrmms^-2\$ in the bottom direction where \$v=0\$. That is, the acceleration is quiet the same.

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But at the highest point, the ball is stationary, so the is not even moving. How can it accelerate?

You litter the ball upwards v velocity \$v\$ and also it returns to your hand with velocity \$-v\$. Let"s attract a graph mirroring the velocity as a duty of time:

Acceleration is identified as:

\$\$ a = fracdvdt \$\$

so that is the gradient of the line in this graph. The velocity-time heat is directly so the gradient is continuous which method the acceleration is constant. The gradient is simply the gravitational acceleration \$9.81\$ m/s\$^2\$.

The allude is that the gradient, and hence the acceleration, does not depend on \$v\$ in ~ all. So that is the same value the \$9.81\$ m/s\$^2\$ once \$v = 0\$ simply as that is at all other values the \$v\$.

When girlfriend shoot the round upwardly, gravity acts upon it with a pressure \$mg\$ wherein \$m\$ is the massive of the ball and also \$g=9.81 ms^-2\$ the Earth"s gravitational acceleration.

If the initial upward velocity to be \$v_0\$ then the instantaneous velocity \$v\$ is given by:

\$v=v_0-gt\$, therefore after part time \$t=fracv_0g\$ the balls"s velocity i do not care \$v=0\$.

However, we know the round will now start falling ago to planet immediately and also if we characterized \$v_0\$ as hopeful then \$v=v_0-gt\$ then now becomes negative. The acceleration \$g\$ hasn"t adjusted though due to the fact that the pressure \$mg\$ acts every the time throughout the trajectory.

The reality that in ~ the apex the its route velocity i do not care momentarily \$0\$ does not mean \$g\$ changes: the doesn"t since the Earth"s gravity acts upon the ball, nevertheless of the velocity or elapsed time.

Acceleration is the price of adjust of velocity. That way it is the distinction of final and also initial velocity divided by the moment duration between these 2 observations. Obviously, that means that there need to be 2 points that time within which acceleration happens. You take any type of two instances that time and get the instantaneous velocities in ~ these 2 instances and also divide that by the expression of time, you room bound to obtain \$9.8 fracms^2\$.

The vital point right here is that while velocity is instantaneous, and therefore have the right to be zero, acceleration is a role of the expression of time, and hence can not be zero.

At the topmost point, the velocity vector is a null vector conversely, the acceleration vector has consistent magnitude \$-9.8,mathrmm/s^2\$ and consistent direction downwards i.e. Towards the center of earth.

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I think you room subconsciously mixing up velocity with acceleration. Let me offer you an example. Imagine these space the measured speed of a bit thrown vertically right into the air at various times:

time, speed0s, 50m/s1s, 40m/s2s, 30m/s3s, 20m/s4s, 10m/s5s, 0m/s6s, -10m/s7s, -20m/s8s, -30m/s9s, -40m/s10s, -50m/s below a minus sign simply means the fragment is coming towards the earth. Currently the acceleration in ~ for example 4s is the difference in between the speeds at 3s and at 5s separated by the elapsed time, which is (20m/s-0m/s)/2s=10m/s^2

Similarly, the acceleration in ~ 5s is the difference in between the speeds at 4s and at 6s separated by the elapsed time, which is (10m/s-(-10m/s))/2s=10m/s^s

This activity is referred to as a movement with consistent acceleration, together the earth exerts a constant force top top the bit at all times, no matter where that is (unless the bit gets really really far, in which instance the force won"t be continuous any more).