Gravity affects the trajectory of every projectile. Gravity is a force and pulls objects down to the Earth. This force is greater on objects with more mass and weaker the farther the objects are from the Earth.

Newton

Newton’s law of gravitation is an inverse square law

where \(G = 6.67 \times 10^{-11} \frac {m^3}{kg \cdot s^2}\) is the gravitational constant.

\(F_g\) is the gravitational force between point-masses \(m_1\) and \(m_2\), which are a distance \(r\) apart.

For us:

• the two masses are our projectile (\(m_p\)) and the Earth (\(m_e\)).
• \(r\) is the distance between the centers of our masses. Our projectiles are a negligent distance off of the Earth. Therefore, we can just say that \(r = r_e\) (the radius of the Earth). \(r\) is basically just a constant for us.

Now, add in the fact that \(F = m \, a\) and this tells us that the gravity(acceleration) a projectile feels from the Earth is given by

A constant!

This is the acceleration we feel on Earth, a downward acceleration, \(-9.81 \frac {m}{s^2}\). That means, for most of our investigations, the acceleration due to the Earth can be considered a constant.

\(\blacktriangleright \) Gravity is constant acceleration.

The acceleration due to the Earth’s gravity is a constant. It is a constant function with respect to time or position.

Acceleration is meters per second PER second. This constant acceleration is a constant rate of change of velocity. Thus, vertical velocity must be a linear function.

This velocity is a linear function. The accumulated vertical distance traveled would be the accumulated area under its graph. We have seen that this area (distance) is a described by a quadratic.

• \(v_0\) is the initial (vertical) velocity for a projectile thrown into the air.
  
• \(s_0\) is the initial height for a projectile thrown into the air.
  
• \(h(t)\) is height (vertical distance) as a function of time. Seconds is the domain unit and meters is the range unit.

Projectile Motion

Actually, all of those quantities are vectors. They are actually holding two pieces of information simultaneously. They are holding a vertical measurement and a horizontal measurement.

Both the vertical, \(y\), and horizontal, \(x\), components of position are functions of time.

We have already seen this equation for the height (vertical distance) of a projectile.

Horizontal motion isn’t effected by the Earth’s gravity, since gravity is a downward acceleration. Therefore, the horizontal distance is just affected by the initial horizontal velocity.

Parabolas

Parabolas are geometric curves. They are defined as the collection of points which are equidistant from a point, called the focus, and a line, called the directrix. In the diagram below, we have rotated our viewpoint so that the directrix is the horizontal line \(y = -c\) and the focus is \((0,c)\).

Projectile Trajectory

We have descriptions of the vertical height and the horizontal distance for a projectile under the influence of gravity. Let’s put those together and get height as a function of distance. That will correspond to our normal experience standing on the ground watching a projectile fly.

• \(h_y(t) = s_{y_0} + v_{y_0} t - \frac {9.81}{2} t^2\)
• \(d_x(t) = s_{x_0} + v_{x_0} t\)

First, let’s go back to our separate equations and let’s just assume we are firing the projectile off the ground. Then our initial position will be \(0\).

And, let’s use the more common symbol \(g\), for the acceleration due to gravity, \(g = \frac {9.81}{2} \frac {m^2}{s}\), downward.

• \(y = v_{y_0} t - g \, t^2\)
• \(x = v_{x_0} t\)

Solve for \(t\) in the horizonal equation.

Substitute this into the vertical equation.

A quadratic!

The projectile itself follows a parabola trajectory in the air.

Separately...

The projectile’s height is described by a quadratic function in time.

The projectile’s horizontal distance is not affected by gravity, so it is described by a linear function in time.