# Why is a projectile a parabola not a semicircle

1. Dec 9, 2015

### Ben Negus

This sounds like a dumb question. I have come to accept projectiles form parabolas but I need someone to explain why they form this shape

2. Dec 9, 2015

### Columbus

Because the force of gravitational acceleration is stronger than the deceleration the projectile experiences. When you view the middle of the flight path, you are dealing with the moments when the vertical velocities are the slowest and horizontal forces appear much more powerful, so that section could be mistaken by the naked eye for a circular path. But the forces are not balanced, so it is parabolic.

3. Dec 10, 2015

### robphy

The parabola is the shape that arises when "the slope of its tangent (which is related to the velocity) changes linearly with time"... [i.e. constant acceleration].

4. Dec 10, 2015

### Staff: Mentor

A projectile is not a parabola, as your thread title states -- the path of a projectile is a parabola in shape, provided that you don't take into account air resistance. The simplistic assumption for determining the path of a projectile is that the horizontal component of its velocity is constant, and that the only force on the projectile is due to gravity. In fact, air resistance affects the horizontal component of velocity, so the range will be less than what is predicted from a parabolic path.

Last edited: Dec 10, 2015
5. Dec 10, 2015

### Svein

To be pedantic, it depends...

If your projectile has enough energy to go into orbit, the path will be a circle (or a descending spiral).

6. Dec 10, 2015

### Staff: Mentor

Good point...

7. Dec 10, 2015

### PeroK

Motion under gravity can be decribed by:

$v_x = vcos(\theta)$ and $v_y = vsin(\theta) - gt$ where $v$ is the projectile's initial velocity at an angle $\theta$ and $g$ is the force of gravity

The $x$ and $y$ coordinates of the motion at time $t$ are given by:

$x = vtcos(\theta)$ and $y = vtsin(\theta) - \frac{1}{2}gt^2$

With a bit of algebra, which I leave to you as an exercise, you can eliminate $t$ to give:

$y = xtan(\theta) - x^2\frac{g}{2v^2}sec^2(\theta)$

This is of the form:

$y = ax - bx^2$ with $a$ and $b$ constants. And that's a parabola.

8. Dec 10, 2015

### 256bits

Neglecting air resistance, isn't the "orbit" of the projectile part of an ellipse.

An actual parabolic path would have to have escape velocity.

The parabolic path of a projectile is actually an approximation.

9. Dec 10, 2015

### Svein

Yes, but good enough for calculating the path of a thrown stone at sea level on Earth.

10. Dec 10, 2015

### DrStupid

c
Under idealized conditions (vacuum, spherical symmetry of the gravitational field) it is in fact an ellipse. For short distances parabolas are very good approximations but in exceptional cases it can also be a circle (circular orbit). In reality it is much more complicate.

11. Dec 10, 2015

### nasu

Regarding your title, even if the trajectory were a segment of circle, it could not be a semicircle. If it were, the initial velocity had to be vertical (tangent to the semicircle). But in this case there will be no horizontal component of the velocity.

12. Dec 10, 2015

### rcgldr

Assuming no atmosphers, then the path of a projectile is really a portion of an ellipse (assuming it's not at or above escape velocity). A parabola is an approximation that treats a gravitational field as constant and uni-directional, such as the field from an infinitely large flat plane. For an object that doesnt' travel significantly (vertically or horizontally) relative to the size of a planet, then a parabola is close enough.