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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?

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    $\begingroup$ Imagine you're in space and your ball has a little rocket engine. You throw the ball away, and the rocket engine is thrusting to return it to you. At the farthest point, before it reverses direction, it briefly stopped moving; but it's clear that the engine is continuously thrusting, accelerating toward you at the same rate. It's because of the constant acceleration that the velocity passes through zero and reverses. $\endgroup$
    – Ask About Monica
    Commented Oct 2, 2015 at 19:04

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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:

Velocity-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 $v$.

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  • $\begingroup$ Thats pure theory. I am taking things a bit practical. @john "But i think, at the very top point, The ball is stationary, it is not moving, How can you say it has a direction?” Thats the question $\endgroup$
    – Aaryan Dewan
    Commented Oct 2, 2015 at 17:14
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    $\begingroup$ @AaryanDewan: The velocity is zero at the top and indeed the velocity has no direction, because a zero vector doesn't have a direction. However the acceleration is non-zero so the acceleration does have a direction. Since $\text{d}\mathbf{v} = \mathbf{a}\text{d}t$ that means the infinitesimal change in the velocity has a direction (the same direction as the acceleration vector). $\endgroup$
    – John Rennie
    Commented Oct 2, 2015 at 17:17
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    $\begingroup$ @AaryanDewan, John Rennie's answer is NOT pure theory. His graph can easily be obtained from experimental data. You seem to be confusing velocity and acceleration, which is a fairly common mistake that I have seen in the class room. $\endgroup$
    – David White
    Commented Aug 12, 2018 at 15:27
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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.

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  • $\begingroup$ But my question is WHY the acceleration is downward when it is stationary? Just because, of The force of gravity ( mg ) ? $\endgroup$
    – Aaryan Dewan
    Commented Oct 2, 2015 at 16:31
  • $\begingroup$ @AaryanDewan: yes, precisely. $g$ is completely independent of $v$. Look at it another way. Your car is stationary at a stop light. Now you push the accelerator and the car starts moving because of the acceleration, say $a$. At $t=0$, $v=0$ and yet acceleration has started. In math terms: $a=\frac{dv}{dt}$, which in no way excludes $t=0,v=0$. An infinitesimal moment $dt$ later $v>0$. $\endgroup$
    – Gert
    Commented Oct 2, 2015 at 16:37
  • $\begingroup$ @AaryanDewan Net force acting on the ball is downwards(mg). So net acceleration is also downwards. THE FORCE mg IS ALWAYS ACTING DOWNWARDS i.e. TOWARDS THE EARTH. $\endgroup$
    – SchrodingersCat
    Commented Oct 2, 2015 at 16:37
  • $\begingroup$ @Gert how can you say, v=0 as you stated in your previous line, that the car started moving? $\endgroup$
    – Aaryan Dewan
    Commented Oct 2, 2015 at 16:41
  • $\begingroup$ @AaryanDewan: because mathematically that is entirely correct. If the acceleration $a$ kicks in at $t=0$ then at that point in time $v=0$. An infinitesimal period of time $dt$ later, at $t=dt$ then $v=0+adt=dv>0$. $\endgroup$
    – Gert
    Commented Oct 2, 2015 at 17:24
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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.

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  • $\begingroup$ Leave the vectors, I am saying that is there is a ball in space, stationary , How can you say it has an acceleration? $\endgroup$
    – Aaryan Dewan
    Commented Oct 2, 2015 at 16:15
  • $\begingroup$ Problem is it is not stationary. Something (read gravity) is pulling it downwards even at that moment. The downward pull actually retarded its upward motion and stopped it and it will take it down. $\endgroup$
    – SchrodingersCat
    Commented Oct 2, 2015 at 16:18
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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).

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  • $\begingroup$ Thats pretty good! But if the earth STILL exerts a force when v=0 and the particle has no direction, how can you say acceleration is downwards? $\endgroup$
    – Aaryan Dewan
    Commented Oct 2, 2015 at 16:37
  • $\begingroup$ Let me put it this way, if you were only given one shot of the motion of the ball, and you didn't know the particle was thrown vertically, there would absolutely be no way of telling what the acceleration is. Acceleration only becomes meaningful when you have at least three shots. Even for the case of two shots, all you can talk about is velocity, and not acceleration. $\endgroup$
    – Mehrdad
    Commented Oct 2, 2015 at 16:45
  • $\begingroup$ what is “shots” ? $\endgroup$
    – Aaryan Dewan
    Commented Oct 2, 2015 at 16:54
  • $\begingroup$ a photograph, a snapshot $\endgroup$
    – Mehrdad
    Commented Oct 2, 2015 at 16:57
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    $\begingroup$ If you take them all at the same time, the ball is stationary, but there is no way to determine either velocity or acceleration. If you take the snapshots at, let's say, 1 millisecond intervals, you'll see that the ball does move between snapshots. So its velocity was zero only for an instant. The velocity reached zero for that instant, but the acceleration was constant. $\endgroup$
    – WhatRoughBeast
    Commented Oct 2, 2015 at 17:19
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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.

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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.

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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.

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  • $\begingroup$ Welcome to Physics Stackexchange. Please go through How to format mathematical formulas? and use the notation scheme wherever necessary. $\endgroup$
    – Yashas
    Commented Jul 31, 2016 at 9:15
  • $\begingroup$ Velocity is the rate of change of position. That means it is the difference of final and initial position divided by the time duration between these two observations. Obviously, that means that there must be two points of time within which velocity happens... The key point here is that while position is instantaneous, and therefore can be zero, velocity is a function of the duration of time, and hence cannot be zero. // See the contradiction yet? Acceleration can be defined as an instantaneous property just as velocity can (thanks, calculus). $\endgroup$
    – electronpusher
    Commented Nov 10, 2021 at 19:14
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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.

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    $\begingroup$ The acceleration $acc=\frac{-u}{t}$ is the average acceleration for the time interval from the when the object is thrown until the point when it is at the top. It's not the instantaneous acceleration at the top (or any other point). $\endgroup$
    – M. Enns
    Commented Jan 9, 2018 at 19:09
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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.

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    $\begingroup$ Welcome to SE. This is not an answer, but a comment. In addition, it's wrong. $\endgroup$
    – FraSchelle
    Commented Nov 22, 2016 at 10:01
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    $\begingroup$ Remember that, whilst on a planet, gravity is always acting on the particle, so it can't be that it 'has zero acceleration at its peak.' $\endgroup$
    – Kyle Kanos
    Commented Nov 23, 2016 at 12:02

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