Acceleration of a ball at its max height after being thrown upwards

[shirozack]

Homework Statement

What is the acceleration of a ball at its max height after being thrown upwards?

Relevant Equations

f=ma

The answer given is -10m/s2 because of constant downwards acceleration of gravity.

i would like to know why is it not 0 at its peak height. at the top, velocity is momentarily 0, since acceleration is the change in velocity, change in 0 = 0. so why issn't a = 0?

thanks

 

[Vanadium 50]

i would like to know why is it not 0 at its peak height. at the top, velocity is momentarily 0, since acceleration is the change in velocity, change in 0 = 0. so why issn't a = 0?

Does it stop at the top?

 

[Orodruin]

Because velocity is only zero for a single moment. Velocity is still changing. Any short time after the apex, the object will be moving down again.

 

[shirozack]

if the ball's velocity is 0 for even a single moment at the top, then the acceleration should be 0 right? v=0, a must be 0?

i understand how gradient of a v-t graph gives acceleration, like if v=0 for 1s, a = (0-0)/1 = 0.

but what if v =0 at only an instantaneous moment? not sure how does the math work out when t->0 ?

 

[PeroK]

if the ball's velocity is 0 for even a single moment at the top, then the acceleration should be 0 right? v=0, a must be 0?

i understand how gradient of a v-t graph gives acceleration, like if v=0 for 1s, a = (0-0)/1 = 0.

but what if v =0 at only an instantaneous moment? not sure how does the math work out when t->0 ?

Newton's second law is $F = ma$, not $F = mv$. The force of gravity doesn't stop acting just because $v = 0$. If it did, then the ball would stop at its highest point and remain at rest there with no force acting on it - as in Newton's first law.

 

[Orodruin]

if the ball's velocity is 0 for even a single moment at the top, then the acceleration should be 0 right? v=0, a must be 0?

A single moment means only at one given time, not an extended time.

i understand how gradient of a v-t graph gives acceleration, like if v=0 for 1s, a = (0-0)/1 = 0.

1 s is not a single moment.

but what if v =0 at only an instantaneous moment? not sure how does the math work out when t->0 ?

Then you have to take the derivative of velocity wrt time. By Newton’s law of gravity as was mentioned above, F = mg leads to a = g.

 

[jbriggs444]

i would like to know why is it not 0 at its peak height. at the top, velocity is momentarily 0, since acceleration is the change in velocity, change in 0 = 0. so why issn't a = 0?

Have you ever been exposed to the mathematical definition for acceleration. Or for the first derivative in general?

The definition does not involve computing the difference between two velocities at the same moment. It involves the limit approached by the difference between two velocities at different moments as the one moment approaches the other.

It is:$$a(t_{top}) = \lim_{t \to t_{top}} \frac{v(t) - v(t_{top})}{t - t_{top}}$$Evaluating that limit does not involve comparing $v(t_{top})$ to itself.

This is a matter of definition. It is what we mean by the word acceleration.

 

[PeroK]

Have you ever been exposed to the mathematical definition for acceleration. Or for the first derivative?

The definition does not involve computing the difference between two velocities at the same moment. It involves the limit approached by the difference between two velocities at different moments as the one moment approaches the other.

It is:$$a(t_{top}) = \lim_{t \to t_{top}} \frac{v(t) - v(t_{top})}{t - t_{top}}$$Evaluating that limit does not involve comparing $v(t_{top})$ to itself.

This is a matter of definition. It is what we mean by the word acceleration.

Also, acceleration is the second derivative of displacement. If we take displacement as a function of time in this case, we have:
$$y = v_0t - \frac 1 2 gt^2$$$$v = \frac{dy}{dt} = v_0 - gt$$$$a = \frac{dv}{dt} = -g$$And we see that the acceleration is constant and does not mysteriously drop to zero just when the velocity happens to be zero.

 

[Orodruin]

It can be added to the above that the same holds for any velocity $v$. The velocity 0 is not special in that sense.

 

[kuruman]

Simply put in layperson's terms, the acceleration of the ball is not zero at maximum height because what goes up must come down.

 

[Steve4Physics]

I'd like a go too. Let's say, using upwards is positive, a ball is thrown with an initial velocity of +20m/s. Take g = -10m/s²
.
The ball's velocity-time graph is this (image from https://socratic.org/questions/a-ba...with-a-vocity-20m-s-it-takes-4-seconds-to-ret ):

The gradient (-10m/s²) is the acceleration; it tells you the rate of change of velocity.

Because we have a straight line, the gradient is the same at every point between t=0 and t=4s.

To find the gradient you could consider the interval from 0 to 4s, or from 0.5s to 3.2s, or from 1.999999s to 2.000001s. You would always get -10m/s².

The gradient at t=2s (when v=0) is the same as at any other time between 0 and 4s.

 

[bob012345]

if the ball's velocity is 0 for even a single moment at the top, then the acceleration should be 0 right? v=0, a must be 0?

i understand how gradient of a v-t graph gives acceleration, like if v=0 for 1s, a = (0-0)/1 = 0.

but what if v =0 at only an instantaneous moment? not sure how does the math work out when t->0 ?

I think I had this issue once way back when I was in school. Kind of like a Zeno’s paradox, if something is nothing how can it change because there’s nothing left to change…

This flawed thinking implies that the velocity itself is what drives the motion rather than gravity. When it runs out there is nothing left. As @Vanadium 50 points out it would stop at the top if that were true. Nature is not like that. Given a constant acceleration, in this case the acceleration due to gravity, we can compute how the velocity constantly changes even as it passes through the $value$ of zero at the top. It would be instructive for you to plot the equations given in post #8.