# Does a thrown ball have kinetic energy at the top of the curve?

I'm going through physics with my 5th grade child. There is a question and answer that indicates that a airborne ball at the top of the trajectory does not have kinetic energy.

1. The diagram below shows the path taken by a ball after it was kicked. The ball hit the ground initially at D and eventually stopped moving at E.

At which position(s) did the ball have no kinetic energy?

1. B only
2. A and E only
3. B and E only
4. B, D, and E only

This is the explanation given in the book:

1. Answer: 3. B and E only

• At A and C, the ball had both kinetic energy and (gravitational) potential energy.
• At the maximum height at B, the ball had only (gravitational) potential energy but no kinetic energy.
• At D, the ball had kinetic energy but no (gravitational) potential energy as it was at the ground level.
• At E, the ball stopped moving, so it had no kinetic energy. The ball also had not (gravitationa) potential energy as it was at ground level.

Ignoring the "complicated" fact that anything with heat has kinetic energy internally, is there some reason the ball wouldn't continue to have kinetic energy? There is no longer vertical motion, but it is still in forward motion.

The answer is wrong. Some author confused the situation when the ball is moving only vertically (and a graph as a function of time) with this case where there is horizontal motion. The horizontal component of the velocity is constant in a ballistic trajectory, it is the same at points A, B, and C.

The kinetic energy is zero only when the ball is stationary, and the ball is stationary only at E: so this is the only point where the kinetic energy is zero.

So... do not trust this book.

The book text is wrong.

If it has kinetic energy at D then it has kinetic energy at B.
There is component of motion in the X direction.

At D there is the potential energy of compressed air (what makes it bounce).

The correct answer E is not even a choice in the book.

Over much deliberation, I agree with Pieter's answer. Perhaps I shall elaborate more. The equation of kinetic energy is $\frac{1}{2}mv^2$. The motion of the ball is a projectile motion, and can be resolved by 2 vectors: the horizontal and the vertical.

The vertical component of this velocity decreases by $-g$ as it approaches the maximum height. Hence, it's velocity at the maximum height is zero, as the energy over here has become potential energy.

The horizontal component, on the other hand, is not zero, and keeps on decreasing, as energy is lost via heat energy.

Hence, our resultant velocity is not zero. Hence $\frac{1}{2}mv^2$ is definitely not zero. Hence, only E, when the ball is at rest, is correct.

First, the picture is really misleading. When I look at it, I seems to indicate that the ball is still moving horizontally, and therefore it has kinetic energy at point B. It wouldn't have kinetic energy at the top only if you throw the ball absolutely straight upwards.

It gets worse. Assuming that the horizontal movement of the drawing just represents a time line and the ball is actually going straight up and down, the answer for point D is very, very dubious. When the ball hits the ground, it starts with kinetic energy due to moving downwards. That energy is converted into potential energy (like a spring) when the ball hits the ground and gets compressed. At the point of maximum compression there is no kinetic energy, the ball stands still. Then it starts expanding again, and the potential energy is again turned into kinetic energy.

• There is a strong indication that the horizontal axis is not time: The axis is a rectangle, representing something like board. – Volker Siegel Jan 14 '18 at 22:27
• The text says "The diagram below shows the path taken by a ball". Emphasis mine. The word "path" indicates that shows the ball in actual space, not in Cartesian coordinates where x is time. – Acccumulation Jan 14 '18 at 23:13
• @Acccumulation "after it was kicked" it's pretty hard to kick it straight it up, is another indicator – Aequitas Jan 15 '18 at 1:54
• @Acccumulation: path doesn't imply anything at all. For example, in thermodynamics, we talk about a path for a series of states through which a system passes from an initial equilibrium state to a final equilibrium state. This path can be represented in P-V, P-T or T-s coordinates. The path of the soccer ball can perfectly well be represented in altitude-time coordinates, not just in x-z. – Eric Duminil Jan 15 '18 at 13:26
• @EricDuminil No, there is a difference between setting up a correspondence between physical space (i.e. points on a piece of paper) and an abstract space, versus a correspondence between physical space and physical space, i.e. a drawing. No axes and no coordinates were given, and it wasn't referred to as a "graph", but a "path". The only reasonable interpretation is that this is a pictorial representation of space. On top of that, it makes no sense to say that the ball stopped moving at E, unless the horizontal direction is spatial. If x were time, then the ball would never stop moving. – Acccumulation Jan 16 '18 at 4:38

As other have stated, the only state where it has no kinetic energy is E, UNLESS the ball is considered to only move in the vertical direction (in which case, the ball has no kinetic energy at B).

As for potential (gravity) energy, if considering the ground as the "absolute" level 0, then it has none, but, obviously, if someone was to dig a hole under the ball, it would fall... which indicates that it DOES have potential energy (but that the energy cannot be converted, as the ball cannot fall), rather than having none at all. However, this "approximation" is being used by most physicists, so we can let it slide.

All in all, the book seems to be garbage in the way it explains things, even if the author seems to know what he's talking about. Better dump it in the trash bin, and look for a better one.

• This post does not serve to add anything, it is a mere repetition. – QuIcKmAtHs Jan 16 '18 at 14:15

If the ball is moving it has kinetic energy. If the ball were kicked straight up, at the apogee its energy would be all potential since it's not moving horizontally or vertically, it stops momentarily. If the ball is kicked in any direction except perfectly vertical, it has kinetic energy throughout its flight since it never stops moving horizontally. The potential energy at any point in its path is gh, where h is how high from a reference, the ground in this case, the ball is at any given time and g, gravity. When the ball is on the ground, PE is zero.

Of course, gravitational potential energy is the kind we're considering vs. stored energy from the kick or bounces.

Yeah, that book stinks!

Good for you doing math homework with your child.

As others have mentioned, the question and answer as written is flawed. The correct answer is of course E Only.

However there is one condition where B could have zero kinetic energy.

That would be if the frame of reference is moving at exactly the same speed as the ball is moving when at the top of the first peak.

That is, if the entire thing above is inside a railway car that accelerates right to that exact speed as the ball reaches B, and you are watching from inside the car.

If the train car then decelerated back to the initial velocity after the ball passed B and reaches E, I guess B and E is possible.

Mind you, the curve for the first loop would look significantly contracted.

Bonus Point: The soccer ball needs air if it bounced like that, or the floor is made of jello.

• The ball still has a velocity in the x direction, i. e. Parallel to the ground, therefore it still has kinetic energy. – physics90 May 8 '18 at 18:30
• @physics90 it's all "relative". Kinetic energy is a relative measure based on the observers frame of reference. Even a ball sitting still on the ground has kinetic energy relative to say the sun. – Trevor_G Jun 8 '18 at 13:55

If it's moving at all, it has kinetic energy, as velocity is a parameter. It is also orbiting around the sun, and would affect any incoming asteroid in it's way, even at rest relative to the Earth at stage E.

The air molecules inside the ball also have kinetic energy, their motion and their collisions with the inside of the ball is what maintains the shape. The ball's surface will also sublimate, even at cold temperatures, as a result of it's random local kinetic energy.

It has less kinetic energy when it is at the top of the arch, this is stored as potential energy relative to the Earth's gravitational field, and re-converted to kinetic energy on the way down. If you threw the ball, directly up, it would have zero motion at the vertex, and the closed chemical system will have no kinetic energy relative to the Earth's gravitational field. If there is any lateral movement at all, it's velocity would make the KE non-zero.

Is it moving?