# Why is the velocity vector perpendicular to the radius vector for projectile motion when the distance from origin is maximal?

After reading this question I have a question of my own:

A solution for the furthest distance from the origin for a projectile motion is to solve $(\frac{y}{x}) \cdot (\frac{dy}{dx})=0$. Why is it that the velocity vector is perpendicular to the radius for a maximum distance?

• "Why is it that the velocity vector is perpendicular to the radius for a maximum distance?" Where did you get that from? – Gert Nov 19 '17 at 21:00
• @Gert : Here is what I mean: physics.stackexchange.com/a/129181/175920. – Joshua Dame Nov 19 '17 at 21:02
• Actually it is $\frac{y}{x}\times \frac{dy}{dx}=-1$ – Gert Nov 19 '17 at 21:16
• @Gert My mistake, I meant: $(\frac{y}{x}) \cdot (\frac{dy}{dx})=0$. However, why does this yield a maximum distance from the origin? – Joshua Dame Nov 19 '17 at 21:25
• I don't think it does (could be wrong of course). I think it's a general condition that applies to all frictionless trajectories. As said, could be wrong on that and I'll eat my hat. ;-) – Gert Nov 19 '17 at 21:34

## 2 Answers

If you have not yet reached maximum distance, then there must be a component of velocity pointing along the radial vector (so the radial vector can still get bigger). Once you have gone past the furthest distance, you must have a component of velocity pointing towards the origin again (otherwise the distance cannot be getting smaller).

It follows that at the point of greatest distance, your velocity has no component along the radius vector: this means it's perpendicular.

Note - this is only true if there is a "true" furthest point - in other words, that there is a point after the furthest point where your distance is smaller again. If you launch a projectile at a low angle, it may never reach this condition (because it hits the ground before reaching the point where the distance starts to decrease again). Therefore this is only applicable when the launch angle is greater than $sin^{-1}\frac89$ as I showed in the linked answer.

First of all, the equation you used in the question is not equivalent to what you intended. The dot product of two vectors is $0$ if and only if the vectors are perpendicular. Also, two slopes in the coordinate plane are perpendicular if and only if their regular (not dot) product is $-1$.

$$\vec v\cdot\vec u=0\Leftrightarrow\vec v\perp\vec u\Leftrightarrow\frac{v_2}{v_1}\times\frac{u_2}{u_1}=-1$$

The question you linked to describes using the expression to find the point where the trajectory is farthest from the object's origin. This trajectory is a smooth curve. Suppose that such a maximum exists. As the object approaches the maximum distance from the origin, the distance is increasing, and the velocity vector of the object has a positive component in the direction away from the origin. That is,

$$\vec r \cdot\vec v>0$$

After the object passes the maximum, the distance is then decreasing, so the velocity vector has a component opposite to the direction from the origin to the object.

$$\vec r \cdot\vec v<0$$

So, the dot product switches from positive to negative as it passes the maximum. Since the velocity components of the object are continuous with respect to time, the dot product is exactly zero at the maximum.

Another intuitive way to observe this is to consider the smallest disk, centered at the origin, that can contain all of the (first quadrant of the) parabola. The edge of the disk then contains the maximum we are considering, since it is the point of the parabola that is the furthest from the origin. Any smaller disk would not contain maximum (and the entire parabola), and any larger disk would not be the smallest possible disk. Again, since the parabola has no sharp corners and must touch the edge of the disk, it must be tangent to the edge of the disk at the maximum. If it were not, the object would go outside the disk just after or before the point in question, contradicting the fact that the parabola is inside the disk. Since the parabola is tangent to the edge of the disk, and the radius of the disk is perpendicular to the edge of the disk, the parabola is perpendicular to the radius at the maximum distance from the origin.

This is an application of the method of Lagrange multipliers. The function we are maximizing is the distance from the origin, and the constraint is the path of the parabola.