As we already know the max range if we launch it from any flat surface or from any height but I was wondering what will be maximum range if we launch it from inclined plane.

I have made on figure of what I observed:

enter image description here

Where alpha is angle from inclined plane and beta be the angle from which the projectile is launched.

  • $\begingroup$ possible duplicate of The trajectory of a projectile launched from a hilltop - although the dupe has the inverse trajectory (throwing downhill instead of uphill) the approach is identical. Note that the answer gives a hint to solving the math - most of the math is given in that question. $\endgroup$
    – Floris
    Commented May 10, 2015 at 19:15

1 Answer 1


This, in fact, is not too different from the flat surface case, the difference lies only in the distribution of the force components along the '$x$ and $y$' components of velocity.

In the flat surface projection case, you have acceleration only along the vertical direction. The given case can be converted to a non-inclined plane case by mentally rotating the system clockwise by $\alpha$ degrees. The only difference this would give you is in the two acceleration components, ($(a_x,a_y)=(-gsin\alpha,-gcos\alpha)$), which implies that both the $x$ and $y$ components of velocity would be affected by the gravitational force, unlike only the vertical component of velocity in the flat surface case. You can obtain a range expression like you did for the flat surface case.

(I've assumed that by range, you mean the distance traveled along the inclined plane.)


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