Projectile motion refers to the motion of an object projected into the air at an angle. Fireworks and water fountains are examples of projectile motion. In this lesson you will learn the fundamentals of projectile motion. It can get confusing so try to keep up or got back and re-read.

The effects of air resistance on the behavior of projectiles can be quite cdifficult to understand. Because effects due to gravity are much simpler and easier to analyze, and since gravity applies in more situations, we will discuss its role in projectile motion. In most instances on Earth, of course, a projectile will be subject to both forces, but there may be specific cases in which an artificial vacuum has been created, which means it will only be subjected to the force of gravity. Furthermore, in outer space, gravity—whether from Earth or another body—is likely to be a factor, whereas air resistance (unless or until astronomers find another planet with air) will not be.

Let us define projectile motion as the motion of a particle through a region of space where it is subject to constant acceleration. An object moving through the air near the surface of the earth is subject to the constant gravitational acceleration, directed downward. If no other forces are acting on the object, i.e. if the object does not have a propulsion system and we neglect air resistance, then the motion of the object is projectile motion.

Do not continue until you have seen an example of the math that is about to be explained, graphically. Click Here or the link at the top of the page.

What follows is a general solution for the two dimensional motion of an object thrown in a gravitational field. The thrown object is called the projectile. Its path is called the trajectory. We will answer all the usual questions that arise in a first year physics class regarding this motion. We will not consider air resistance. Without air resistance, the projectile will follow a parabolic trajectory. We will be throwing the projectile on level ground on planet Earth. It will leave the point of release, arc through the air along a path shaped like a parabola, and then hit ground a certain distance from where it was thrown.

As mentioned above, this is a two dimensional problem. Therefore, we will consider x and y directed displacements, velocities, and accelerations. The projectile will accelerate under the influence of gravity, so its y acceleration will be downward, or negative, and will be equal in size to the acceleration due to gravity on Earth. There will be no acceleration in the x direction since the force of gravity does not act along this axis.

On Earth the acceleration due to gravity is 9.8 m/s² directed downward. So, for this presentation acceleration in the y direction, or a_y, will be -9.8 m/s², and acceleration in the x direction, or a_x, will be 0.0 m/s².

The original conditions are the size of the velocity and the angle above the horizontal with which the projectile is thrown. The original size of velocity is "v_o". The original angle is set to "theta". Now let us set the actual values. Let v_o = 40.0 m/s. Let theta = 35 degrees.

The first step in this investigation is to find the x and y components for the original velocity. Let us set the X component of original velocity: v_ox = v_ocos(theta). Also, let the Y component of original velocity: v_oy = v_osin(theta). When we do the math with the examples from above the X component is v_ox = 32.8 m/s. When we do the math with the examples from above the Y component is v_oy = 22.9 m/s.

At the top of the trajectory the y, or upward, velocity of the projectile will be 0.0 m/s. The object is still moving at this moment, but its velocity is purely horizontal. At the top it is not moving up or down, only across.

Notice that the object is still in motion at the top of the trajectory; however, its velocity is completely horizontal. It has stopped going up and is about to begin going down. Therefore, its y velocity is 0.0 m/s.

We need to find out how much time passes from the time of the throw until the time when the y velocity of the projectile becomes 0.0 m/s. This y velocity at the top of the trajectory can be thought of as the final y velocity for the projectile for the portion of its flight that starts at the throw and ends at the top of the trajectory.

We will call this amount of time 'the half time of flight', since the projectile will spend one half of its time of flight rising to the top of its trajectory. It will spend the second half of its time of flight moving downward.

t = (v_fy - v_oy) / a_y

t = 2.33 s

Now we need to find the displacement in the y direction at the time when the projectile is at the top of its flight. We have just found the time at which the projectile is at the top of its flight. If we plug this time into a kinematics formula that will return the displacement, then we will know how high above ground the projectile is at when it is at the top of its trajectory.

d_y = v_oyt + 0.5a_yt²

d_y = 27 m