Today, I was learning about the time of flight of a parabola on an incline plane, which I wanted to derive $$T = \frac{2v_0 \sin \theta}{g \cos \alpha}$$ where $\theta$ is the angle of projection w.r.t to the incline and $\alpha$ is the angle of the inclined plane.

Image of Projectile

Image from [Projectile on inclined plane, angle for maximum range]

First, I made the $x$ axis parallel to the inclined plane and the $y$ axis perpendicular to it.

Now, I thought that the projectile was just a simple parabola, so I thought about computing the time to reach the peak and doubling it to get the time of flight.

Time to reach peak can be computed from $$0 = v_0 \sin \theta - gt \cos \alpha$$ as $$t = \frac{v_0sin\theta}{g\cos\alpha}.$$

Doubling it gives the desired. However, from the images I've seen online, it does not seem to me that the projectile is simple parabola (as in the ground case) that can be "cut" into halves. So, why does this work?

The image below is the standard ground to ground parabola. Notice how the motion can be "divided" into two analogous parts.

Ground to ground parabola

I am sorry if I am missing some trivial details.

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    $\begingroup$ In this context the word "peak" means maximum distance from the plane measured normal to the plane. It not not mean the maximum height measured from the horizontal. $\endgroup$ Jul 16, 2021 at 4:45
  • 3
    $\begingroup$ Technically, its still a parabola, just not a standard one. Rotation transform doesn't change the nature of the conic section. good Q btw. $\endgroup$
    – lineage
    Jul 16, 2021 at 16:38
  • $\begingroup$ Your "peak" is maximum vertical distance to the inclined plane, not maximum orthogonal distance. This is why it works. $\endgroup$ Jul 16, 2021 at 23:37
  • $\begingroup$ If you are confused about why the parabola along the slope isn't symmetrical its because gcosα continually decreases the horizontal velocity along the slope. So thats why the parabola looks squished here. $\endgroup$ Sep 26, 2022 at 19:34

7 Answers 7


You have asked a good question.

…it does not seem to me that the projectile is simple parabola (as in the ground case) that can be "cut" into halves. So, why does this work?

Really it contrasts with the symmetry. But it has nothing to do with symmetry. You will better understand with following graphs.


Here black line indicates the trajectory of the ball and brown line is the inclined plane.

As @John Rennie precisely explained,

....the word "peak" means maximum distance from the plane measured normal to the plane. It does not mean the maximum height measured from the horizontal. 

I have marked them as 'A' and 'B' respectively.


If you try graphing this yourself you will realize that distances $a$ and $b$ in the following picture are equal (This property was discovered by Archimedes).*


Since there is no horizontal acceleration, the amounts of time that the ball takes to pass $a$ and $b$ are equal (consider horizontal motion). Eventually you will conclude that that the time of the flight is given by doubling the time to reach the peak(A).

Thus it is not due to symmetry, but because of the special property of intersection of a straight line and a parabola.

Hope this helps.

P.S.: @Fredriksy has also explained the same thing in his answer,

It seems like you are worried that the flight path after you rotate the picture (x-y axes) is NOT a parabola. However this is not actually important for determining the time of flight.

I guess, with this explanation and my graphs you will understand better. Good luck.

*You can find the mathematical proof here.
(Special thanks go to @CiaPan and @Pope)


Can you observe something else interesting? If you consider a projection relative to the horizontal plane, the horizontal plane will also be a chord to the trajectory, which is a parabola. So the observation, 'by doubling the time it takes to reach maximum altitude with respect to the considered plane, flight time is obtained ', can also be interpreted as a result of this special property of intersection, although it obviously seems to be a consequence due to symmetry:-)


I think you have a misunderstanding as to why your solution works.

You are correct in that a parabolic path will “peak” at the top and will take equally long to drop down. However this is not restricted to just the vertically symmetric parabolic path.

In you case you have correctly separated out the x and y velocities and accelerations as $$v_y(t=0) = v_0\sin\theta$$ $$a_y = -g\cos\alpha$$

$$v_x(t=0) = v_0\cos\theta$$ $$a_x = -g\sin\alpha$$

It seems like you are worried that the flight path after you rotate the picture (x-y axes) is NOT a vertically symmetric parabola. However this is not actually important for determining the time of flight. The only thing that matters is $v_y$ And $a_y$ In your rotated picture. The x direction $v_x$ And $a_x$ that make your flight path not symmetrical/parabolic doesn’t affect the time of flight, it will only affect where it lands. I’ll say it again , the path does not need to be symmetric or parabolic such that the time it takes to reach the top is equal to the time it takes to reach the ground.

So the question then becomes, why is $t_{\text{up}}=t_{\text{down}}?$

Argument: $$ t_{\text{up}} = -v_y(t=0)/a_y \; \text{time it takes to get to zero velocity at the top} $$ $$ t_{\text{down}} = -v_y(t=0)/a_y \; \text{} $$

this time we know, from symmetry in the $y$ motion between the upward and downward phases (uniformly accelerated $y$ motion), that the $v_y(t=\text{Return})$ when it comes down is $-v_y(t=0)$. So we want to know the time it takes to go from $0$ velocity (the top) to $-v_y(t = 0)$ when hitting the ground.

Another example is that if there is wind blowing in the x direction, if you throw the ball in the air it will peak and then drop down. $t_{\text{up}} = t_{\text{down}}$ , even though the ball will drop somewhere else entirely due to the wind.

Edit: added some $t = 0$ notes to clarify things

  • $\begingroup$ 1. It should be $a_y = -g\cos\alpha$ and $a_x = -g\sin\alpha$. 2. Any rotation of a parabola is a parabola. It's just not symmetric around the $y$ axis. 3. "we know from conservation of energy" -- I don't think that works. The kinetic energy upon landing is not the same (due to the change in potential energy), and $v_x$ is not the same due to $a_x$. Rather, the argument should be that the same $y$ distance is traveled with uniform acceleration $a_y$ in equal times before and after the time when $v_y = 0$ (as in John Hunter's answer). $\endgroup$
    – nanoman
    Jul 17, 2021 at 6:22
  • $\begingroup$ 4. The wind analogy is questionable. You'd need to take the limit of a very strong horizontal wind and a very dense projectile, for drag to affect the horizontal motion but not the vertical motion. $\endgroup$
    – nanoman
    Jul 17, 2021 at 6:50
  • $\begingroup$ The comment on conservation of energy argument is correct , feel free to post a correction. The horizontal wind analogy seems fine to me, it’s just a simple model after all. $\endgroup$
    – Fredrik Sy
    Jul 17, 2021 at 7:05

Your method works ok and the reason is this.

For a projectile fired over horizontal ground, the time of flight is double the time taken for it to get to the highest point.

This time can be found by considering vertical components only - it is the same time that a projectile fired vertically upwards would take to go up and down - i.e. horizontal components of velocity can be ignored.

When you rotate the axes it's like a projectile fired at an angle $\theta$ to horizontal ground, with $g$ reduced to $g\cos\alpha$ but a 'horizontal' acceleration $g\sin\alpha$ added to the projectile.

However this 'horizontal' component doesn't affect the time for it to go to different distances above the plane, so the time to reach the maximum distance is again half the time of the full motion.

  • 2
    $\begingroup$ Nice answer. Note that the time to the "peak" that the OP found would not actually be the highest point vertically. Instead, it would be the time at which the component of the velocity perpendicular to the inclined plane is zero; or, to put it another way, the point at which the velocity is parallel to the inclined plane. $\endgroup$ Jul 16, 2021 at 11:47
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    $\begingroup$ @Michael Seifert , thanks, by 'peak' yes, the OP means the point farthest from the plane. $\endgroup$ Jul 16, 2021 at 16:15

Note that if you drew a line perpendicular to the incline plane where the projectile was launched, and a line perpendicular to the incline plane where the projectile landed, then a distance line drawn in between these two lines parallel to the plane, defines the entire time of flight of the projectile (up to where it lands back on the incline), say $t$.

The "highest point" of the projectile corresponds to the greatest length line measured perpendicular to the incline from the incline to the path of the projectile. In other words, a tangent at that point is on the parabola is parallel to the incline. This corresponds to time $\frac{t}{2}$ since the x-y plane has been rotated by $\alpha$, and the acceleration $g$ is now $gcos\alpha$ with an acceleration $gsin\alpha$ parallel to the incline.

Note that if you removed the incline, the path taken by the projectile would resemble a perfect parabola, and the symmetry that you would expect, would appear, and hence why you would think that.

  • $\begingroup$ I don't understand this answer -- there is an acceleration component parallel to the inclined plane, so half the time does not correspond to half the distance. $\endgroup$
    – nanoman
    Jul 17, 2021 at 6:08
  • $\begingroup$ Have a read of ACB’s answer. He has diagrams to help visualise the situation. Cheers. $\endgroup$
    – joseph h
    Jul 17, 2021 at 8:18
  • $\begingroup$ What is in ACB's answer doesn't change my criticism of this answer. You say "half-way along this line...determines half the time", but you don't explain exactly what it means for the projectile to be "half-way along this line", or why half-way is when the projectile reaches maximum distance from the inclined plane. $\endgroup$
    – nanoman
    Jul 17, 2021 at 8:32
  • $\begingroup$ If I get a little time later, I'll expand on my answer. $\endgroup$
    – joseph h
    Jul 17, 2021 at 8:34
  • $\begingroup$ Alas, I still don't see a clear argument here. You say "the greatest length line measured perpendicular to the incline from the incline to the path of the projectile, corresponding to time $t/2$", but don't say why this greatest distance corresponds to time $t/2$. I think what you're driving at is that you can separate the equations for motion parallel to and perpendicular to the incline, and analyze only the latter, which then reduces to ordinary projectile motion (with gravity $g\cos\alpha$). $\endgroup$
    – nanoman
    Jul 17, 2021 at 22:19

It's the same as introducing an acceleration in horizontal direction for a thrown ball on a horizontal plane. The inclined plane be coordinated so that the inclined line is the x-axis and a line orthogonal to it the y-axis. There is a an acceleration in the x-direction too. This will not alter the height of the parabola's top. Only its x-position. Vertical gravity is reduced in this coordinate syste. The ball will still reach its top in half the time as without this new horizontal force. The (parabolic) trajectory of the ball will be unevenly squeezed or elongated ( when thrown from the top of the inclined plane). The time to reach the top and the time from the top till touchdown will not be altered. Only the horizontal motion will be different (non-linear)


We want to maximize $ \vec{s} \cdot \hat{n}$ where $n$ is the unit normal to the ramp. $$ \vec{s} = \vec{u} t + \frac12 \vec{a} t^2$$

Dot both sides with $\hat{n}$:

$$ \vec{s} \cdot \hat{n} = \vec{u} \cdot \hat{n} t + \frac12 \vec{a} \cdot \hat{n} t^2$$

Differentiate both sides with time:

$$ \frac{d}{dt}( \vec{s} \cdot \hat{n}) = \vec{u} \cdot \hat{n} + \vec{a} \cdot \hat{n} t$$

For the maximum case, it must be that $\frac{d}{dt} ( \vec{s} \cdot \hat{n})=0$, hence by rearrangement:

$$\left| \frac{\vec{u} \cdot \hat{n} }{ \vec{a} \cdot \hat{n}}\right|= t$$

We find $\vec{u} \cdot \hat{n} = |\vec{u}| \sin \theta$ and $ \vec{a} \cdot \hat{n}=| \vec{a} | \cos \alpha$ (by geometry):

$$ t=\left| \frac{|\vec{u}_0| \sin \theta}{|\vec{a}| \cos \alpha}\right|$$

Beauty of the above proof: nowhere did I use coordinates or components!

  • $\begingroup$ This doesn't seem to answer the question. You compute the time of maximum distance from the inclined plane, but don't show that the time of landing is twice that. $\endgroup$
    – nanoman
    Jul 17, 2021 at 6:11

The distance from the inclined plane does not have to measured normal to the plane. It can be measured at any fixed angle, and the point of maximum distance will be the same. In particular, using the original $y$ and $z$ coordinates in OP's diagram, we can define the peak using the vertical distance $$Z = z - y\tan\alpha,$$ i.e., the difference between $z$ and the height of the inclined plane at the same $y$ value.

The maximum value of $Z$ occurs when $0 = dZ/dy = dz/dy - \tan\alpha$, i.e., when the trajectory has slope $\tan\alpha$ and has a tangent line parallel to the inclined plane. This is exactly the same as the criterion when measuring distance normal to the plane.

We can view the entire problem in the coordinates $(y,Z)$, which are an affine shear transformation of the original ones. We have $dy/dt = \mathrm{const}$ and $d^2Z/dt^2 = d^2z/dt^2 = -g$. So projectile motion in $(y,Z)$ follows parabolic trajectories of exactly the same kind as in $(y,z)$. The transformation turns the problem into one with level ground. This makes it clear that the relation between peak time (maximum $Z$) and landing time ($Z = 0$) is the same as in the level-ground case.


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