# Maximum velocity for projectile motion?

• B
• Feynstein100
In summary: No, the object never reaches a maximum or minimum. The maximum( absolute(velocity) ) will be when the height, y = 0.
Feynstein100
Hey so I encountered a problem recently, which while being simple, gave me another problem that I couldn't solve.
The problem:
An object is launched vertically upward with an initial velocity of 10 m/s from an elevation of 20m and allowed to hit the ground. Develop the equations of motion between elevation & velocity vs time. Also find the maximum elevation and maximum velocity attained by the object.
The first part was too trivial to mention here. It was the second part that got me.
The maximum elevation was quite simple. I had an equation for elevation vs time. All I had to do was differentiate it and set it to zero. I noticed that this was intuitive because the derivative of elevation would be velocity and since this is projectile motion i.e. constant acceleration, its velocity at the highest point would be zero. And thus I was able to find the max elevation.
However, when it came to max velocity, I thought I could simply do the same thing again, since I had an equation for v vs t. But then I realized that if I differentiate that equation, I get acceleration, which is constant and not equal to zero.
This means I can't use the maxima/minima property of derivatives to find max velocity. Why didn't that work?
Okay I see now that the velocity increases linearly, meaning its graph on v vs t is a straight line. As such, its maxima and minima are at +
and -
respectively. Which is probably why it didn't work.
Is this only true for straight lines or are there other curves where the maxima-minima property doesn't work? Or at least, isn't useful

Feynstein100 said:
Hey so I encountered a problem recently, which while being simple, gave me another problem that I couldn't solve.
The problem:
An object is launched vertically upward with an initial velocity of 10 m/s from an elevation of 20m and allowed to hit the ground. Develop the equations of motion between elevation & velocity vs time. Also find the maximum elevation and maximum velocity attained by the object.
The first part was too trivial to mention here. It was the second part that got me.
The maximum elevation was quite simple. I had an equation for elevation vs time. All I had to do was differentiate it and set it to zero. I noticed that this was intuitive because the derivative of elevation would be velocity and since this is projectile motion i.e. constant acceleration, its velocity at the highest point would be zero. And thus I was able to find the max elevation.
However, when it came to max velocity, I thought I could simply do the same thing again, since I had an equation for v vs t. But then I realized that if I differentiate that equation, I get acceleration, which is constant and not equal to zero.
This means I can't use the maxima/minima property of derivatives to find max velocity. Why didn't that work?
Okay I see now that the velocity increases linearly, meaning its graph on v vs t is a straight line. As such, its maxima and minima are at +
and -
respectively. Which is probably why it didn't work.
Is this only true for straight lines or are there other curves where the maxima-minima property doesn't work? Or at least, isn't useful
Where the velocity is maximum should be intuitive as well if only conservative forces are acting on the projectile.

@erobz is correct, but you are also missing part of the maximum/minimum properties of derivatives. What do you do to check that you have not found a local extremum for the general case?

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PeroK, Feynstein100 and erobz
Feynstein100 said:
This means I can't use the maxima/minima property of derivatives to find max velocity. Why didn't that work?
Because the object hits the ground and so the function is discontinuous.
The maximum( absolute(velocity) ) will be when the height, y = 0.

Baluncore said:
Because the object hits the ground and so the function is discontinuous.
The maximum( absolute(velocity) ) will be when the height, y = 0.
Since the answer was given, I’ll give the why’s
1) Mathematically, you also have to check the boundaries which are t=0 and when the projectile hits the ground.
2) Physically the potential energy is lowest when the projectile hits the ground so the kinetic energy (i.e. velocity) is at its maximum.

erobz said:
Where the velocity is maximum should be intuitive as well if only conservative forces are acting on the projectile.
It was, but that's not the point of my question

Baluncore said:
Because the object hits the ground and so the function is discontinuous.
The maximum( absolute(velocity) ) will be when the height, y = 0.
For this case, yes. But imagine the line y = x. It is indeed continuous everywhere but when you calculate the derivative, dy/dx = 1, which can't be set to zero. That's kind of what I was asking. Does this function not have a maxima/minima?

Frabjous said:
1) Mathematically, you also have to check the boundaries which are t=0 and when the projectile hits the ground.
I don't understand. What do you mean?

Feynstein100 said:
I don't understand. What do you mean?
In general, when calculating extremum you also have to check the values at the boundaries. For this problem, the boundaries are when you launch the object and when the object hits the ground. The equations that you developed are invalid outside this region.

berkeman
Compare the energy at each condition: Initial , apogee, final. It should be constant ( between resting)

Feynstein100 said:
Does this function not have a maxima/minima?
Where y = x, without bounds, there is no turning point.
dy/dx = 1 = constant.

Feynstein100 said:
For this case, yes. But imagine the line y = x. It is indeed continuous everywhere but when you calculate the derivative, dy/dx = 1, which can't be set to zero. That's kind of what I was asking. Does this function not have a maxima/minima?
Yes, at the end points. Points with zero derivative are not the only possibility for maxima and minima.

renormalize
PeroK said:
Yes, at the end points. Points with zero derivative are not the only possibility for maxima and minima.
I thought that was the definition of maxima and minima

Baluncore said:
Where y = x, without bounds, there is no turning point.
dy/dx = 1 = constant.
I noticed that too. It looks like the derivative is only zero when there's a turning point. Not all functions have them

Feynstein100 said:
I thought that was the definition of maxima and minima
##dy(x)/dx=0## is only the definition for local extrema. There can be global maxima or minima at the endpoints of the interval of interest.

renormalize said:
##dy(x)/dx=0## is only the definition for local extrema. There can be global maxima or minima at the endpoints of the interval of interest.
Huh. I never learned that in calculus class

Feynstein100 said:
Huh. I never learned that in calculus class
He just means you could have a function that has a maximum or minimum value but the slope isn't zero.

Your real problem here is that you are not looking at the horizontal and vertical components of the position and velocity. It is those functions that have the properties of maxima and minima that you seek.

gmax137
renormalize said:
##dy(x)/dx=0## is only the definition for local extrema. There can be global maxima or minima at the endpoints of the interval of interest.
In general, you also have to look for discontinuities and points where the function is not differentiable. For example, there is a local and global minimum of the modulus function at ##0##, where the derivative does not exist.

SammyS and Ibix
Feynstein100 said:
I thought that was the definition of maxima and minima
The maxima and minina are defined to be the points where a function attains its greatest or least value. This definition applies to all functions, whether differentiable or not.

A.T. and jbriggs444
Part of the problem here is that you aren't modelling the velocity correctly. It's not ##\vec{v}=(v_{x0},v_{y0}-gt)^T##, it's $$\vec{v}=\left\{\begin{array}{ll}(v_{x0},v_{y0}-gt)^T&0\leq t\leq T\\(0,0)^T&\mathrm{otherwise}\end{array}\right.$$where ##T## is the flight time and the subscripted ##v## are initial velocity components. That's because your projectile was stationary before launch and becomes stationary again when it hits the ground. So your vertical v/t graph isn't just a straight line, it's actually more like this sketch:

Simply taking the derivative of ##v_{y0}-gt## ignores the discontinuities and assumes the projectile is in motion forever. But there are actually three phases of its motion with discontinuities in the velocity between them. And those, as others have commented, you have to check for separately from your calculus.

DrClaude
Feynstein100 said:
Huh. I never learned that in calculus class
You probably were taught it, its just one of those things that are said in passing on the way to the interesting part.

For instance, over the interval ##a## to ##b## the critical points of this function (what you find from ##y'(x)=0##) are not the largest (or smallest) values the function assumes on the interval.

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DrClaude and renormalize
PeroK said:
The maxima and minina are defined to be the points where a function attains its greatest or least value. This definition applies to all functions, whether differentiable or not.
More formally, a point ##x## is a "local maximum" of a function ##f## if and only if there is some finite neighborhood around ##x## where the function attains no value greater than ##f(x)##.

Or more formally yet:

##f## is defined at ##x## and there exists a real number epsilon (##\epsilon##) greater than zero such that for every point ##y## in the open interval from ##(x - \epsilon, x + \epsilon)## that also falls within the domain of ##f##, ##f(y) \le f(x)##.A point ##x## is a "maximum" or "global maximum" if there is no point anywhere within the domain of the function where the function attains a value greater than ##f(x)##.Given the above definitions, we can prove the following:

If a point ##x## is a local extremum of a function ##f## then one of the following must be the case:

1. ##f## is not differentiable at ##x##.

E.g. ##f(x) = |x|## with a minimum at ##x = 0##, but ##f## has no first derivative at ##x = 0##. This function has no maximum.

2. ##f'(x) = 0##

E.g. ##f(x) = x^2## with a minimum at ##x = 0##
E.g. ##f(x) = 1##. A degenerate case with a global maximum and global minimum everywhere

3. ##x## is a boundary point in the domain of ##f##

E.g. ##f(x) = x## for ##0 \le x \le 1## with a minimum at ##x = 0## and a maximum at ##x = 1##

PeroK
Feynstein100 said:
Huh. I never learned that in calculus class
Do you need calculus to understand what "maximum" or "minimum" means?

Feynstein100 said:
I thought that was the definition of maxima and minima
Why would you think that? A saddle point has zero slope, but is neither a maximum nor minimum.

Zero slope is a required condition for local maximum or minimum of a continuous function, but it doesn't imply one.

As a more-or-less non-mathematician and non-physicist, I would suggest that it is simply the difference between English and Maths. Maximum and minimum have perfectly well-known and understood meaning in English. They also have a definition in Maths (I presume!) Maths is sometimes helpful in solving problems explained in English, but such problems do want common sense and realistic answers, whatever maths says.

The other place where this always used to arise for basic projectile kinematics problems, is in the solution of quadratic equations giving rise to two answers for when the projectile (from the top of a building, for example) hits the ground. One often had to explain the existence of a mathematical solution that had it hitting the ground before you threw it.

A.T. said:
Zero slope is a required condition for local maximum or minimum of a continuous function, but it doesn't imply one.
Be careful. "Continuous" does not mean what you think it means.

You probably want "continuously differentiable over an open interval containing the extremum in question" or, less restrictive, "differentiable at the extremum such that both left and right side derivatives exist and are equal".

1. A function is continuous at every isolated point in its domain. It will not have a defined slope there. But the isolated point is sure to be both a local minimum and a local maximum.

e.g. ##f(x) = 1## for ##x = 1## with ##f## undefined elsewhere​

2. A function may be continuous but have a non-zero slope at an end point in its domain. Yet the end point may be a local extremum.

e.g. ##f(x) = x## for ##0 \le x \le 1## with ##f## undefined elsewhere​

3. A function may be continuously defined over an open interval containing a point with a local extremum but fail to have a defined slope at that point.

While it is tempting to invoke the Weirstrass function (something of a universal counterexample), a more mundane example will suffice​

e.g. ##f(x) = |x|## at ##x=0##​

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A.T.
jbriggs444 said:
Be careful. "Continuous" does not mean what you think it means.

You probably want "continuously differentiable over an open interval containing the extremum in question" or, less restrictive, "differentiable at the extremum such that both left and right side derivatives exist and are equal".
Yes, I got used to just talk about "continuity" (of certain order) rather than "differentiabilty" (of one order less):

https://en.wikipedia.org/wiki/Smoothness#Order_of_parametric_continuity

Should have been more clear that I mean C1.

jbriggs444

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