A projectile is an object upon which the only force acting is gravity. Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. That is, as they move upward and/or downward they are also moving horizontally. There are the two components of the projectile's motion - horizontal and vertical motion. And since "perpendicular components of motion are independent of each other," these two components of motion can (and must) be discussed separately. The goal of this part of the lesson is to discuss the horizontal and vertical components of a projectile's motion; specific attention will be given to the presence/absence of forces, accelerations, and velocity.
Consider a cannonball projected horizontally from a cannon from upon a very high cliff. In the absence of gravity, the cannonball would continue its horizontal motion at a constant velocity. This is consistent with the law of inertia. And furthermore, if merely dropped from rest in the presence of gravity, the cannonball would accelerate downward, gaining speed by a rate of 10 m/s every second. This is consistent with our conception of free-falling objects accelerating at a rate we call the "acceleration of gravity."
If our thought experiment continues and we project the cannonball horizontally in the presence of gravity, then the cannonball would maintain the same horizontal motion as before - a constant horizontal velocity. Furthermore, the force of gravity will act upon the cannonball to cause the same vertical motion as before - a downward acceleration. The cannonball falls the same amount of distance as it did when it was merely dropped from rest (refer to diagram). However, the presence of gravity does not effect the horizontal motion of the projectile. The force of gravity acts downward and is unable to alter the horizontal motion. There must be a horizontal force to cause a horizontal acceleration. (And we know that there is only a vertical force acting upon projectiles.) The vertical force acts perpendicular to the horizontal motion and will not effect it ("perpendicular components of motion are independent of each other"). Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.
The above information can be summarized by the following table.
| | Horizontal Motion | Vertical Motion |
| Forces
(Present? - Yes or No) If present, what dir'n) | No | Yes The force of gravity acts downward |
| Acceleration
(Present? - Yes or No)
If present, what dir'n) | No | Yes "g" is downward at ~10 m/s/s |
| Velocity
(Constant or Changing) | Constant | Changing (by ~10 m/s each second) |
In this portion of Lesson you will learn how to describe the motion of projectiles numerically. You will learn how the numerical values of the x- and y-components of the velocity and displacement change with time (or remain constant). As you proceed through this part of Lesson , pay careful attention to how a conceptual understanding of projectiles translates into a numerical understanding.
Consider again the cannonball launched by a cannon from the top of a very high cliff. Suppose that the cannonball is launched horizontally (with no upward angle whatsoever) with an initial speed of 20 m/s. If there were no gravity, the cannonball would continue in motion at 20 m/s in the horizontal direction. Yet in actuality, gravity causes the cannonball to accelerate downwards at a rate of ~10 m/s/s. This means that the vertical velocity is changing by ~10 m/s every second. If a vector diagram (showing the velocity of the cannonball at 1-second intervals of time) is used to represent how the x- and y-components of the velocity of the cannonball is changing with time, then x- and y- velocity vectors could be drawn and their magnitudes labeled. The length of the vector arrows are representative of the magnitudes of that quantity.
The important concept depicted in the above vector diagram is that the horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by ~10 m/s every second. These same two concepts could be depicted by a table illustrating how the x- and y-component of the velocity vary with time.
| Time | Horizontal Velocity | Vertical Velocity |
| 0 s | 20 m/s, right | 0 |
| 1 s | 20 m/s, right | 10 m/s, down |
| 2 s | 20 m/s, right | 20 m/s, down |
| 3 s | 20 m/s, right | 30 m/s, down |
| 4 s | 20 m/s, right | 40 m/s, down |
| 5 s | 20 m/s, right | 50 m/s, down |
The numerical information in both the diagram and the table above illustrate identical points - a projectile has a vertical acceleration of ~10 m/s/s, downward and no horizontal acceleration. This is to say that the vertical velocity changes by ~10 m/s each second and the horizontal velocity never changes. This is indeed consistent with the fact that there is a vertical force acting upon a projectile but no horizontal force. A vertical force causes a vertical acceleration - in this case, an acceleration of ~10 m/s/s.
Describing Projectiles With Numbers
(Horizontal and Vertical Displacement)
The previous diagrams, tables, and discussion pertain to how the horizontal and vertical components of the velocity vector change with time during the course of projectile's trajectory. Now we will investigate the manner in which the horizontal and vertical components of a projectile's displacement vary with time. As has already been discussed, the vertical displacement (denoted by the symbol y in the discussion below) of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity. Thus, the vertical displacement (y) of a projectile can be predicted using the same equation used to find the displacement of a free-falling object undergoing one-dimensional motion. The equation can be written as follows.
y = 0.5*g*t2
(eq'n for vertical displacement for a horizontally-launched projectile)
where g = -10 m/s/s (approximated value for the acceleration of gravity) and t = time (in seconds). The above equation pertains to a projectile with no initial vertical velocity and as such predicts the vertical distance which a projectile falls if dropped from rest. It was also discussed earlier, that the horizontal motion of a projectile is not influenced by the force of gravity. The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally and the amount of time (t) which it has been moving horizontally. Thus, if the horizontal displacement (x) of a projectile were represented by an equation, then that equation would be written as
x = vix * t
The diagram below shows the trajectory of a projectile (in red), the path of a projectile released from rest with no horizontal velocity (in blue) and the path of the same object when gravity is turned off (in green). The position of the object at 1-second intervals is shown. In this example, the initial horizontal velocity is 25 m/s and there is no initial vertical velocity (i.e., a case of a horizontally-launched projectile).
The vertical distance fallen from rest during each consecutive second is increasing (i.e., there is a vertical acceleration). It can also be seen that the vertical displacement (y) follows the equation above (y=.5*g*t2). Furthermore, since there is no horizontal acceleration, the horizontal distance traveled by the projectile each second is a constant value - the projectile travels a horizontal distance of 25 meters each second. This is consistent with the initial horizontal velocity of 25 m/s. Thus, the horizontal displacement is 25 m at 1 second, 50 meters at 2 seconds, 75 meters at 3 seconds, etc. This information is summarized in the table below.
| Time | Horizontal Displacement | Vertical Displacement |
| 0 s | 0 m | 0 m |
| 1 s | 25 m | 5 m |
| 2 s | 50 m | 20 m |
| 3 s | 75 m | 45 m |
| 4 s | 100 m | 80 m |
| 5 s | 125 m | 125 m |
Now consider displacement values for a projectile launched at an angle to the horizontal (i.e., a non-horizontally launched projectile). How will the presence of a initial vertical component of velocity effect the values for the displacement? The diagram below depicts the position of a projectile launched at an angle to the horizontal. The projectile still falls 5 m, 20 m, 45 m, and 80 m below the straight-line, gravity-free path.
The projectile still falls below its gravity-free path by a vertical distance of 0.5*g*t^2. However, the gravity-free path is no longer a horizontal line since the projectile is not launched horizontally. In the absence of gravity, a projectile would rise a vertical distance equivalent to the time multiplied by the vertical component of the initial velocity (viy*t). In the presence of gravity, it will fall a distance of 0.5*g*t^2. Combining these two influences upon the vertical displacement yields the following equation.
y = viy * t + 0.5*g*t2
where viy=initial vertical velocity (in m/s), t=time (in seconds), and g=-10 m/s/s (the acceleration of gravity). If a projectile is launched with an initial vertical velocity of 20 m/s and an initial horizontal velocity of 34.6 m/s, then the x- and y- displacements of the projectile can be calculated using the equations above. A sample calculation is shown below.
| Calculations for t = 1 second | y = viy * t + 0.5*g*t2 where viy = 20 m/s y = (20 m/s) * (1 s) + 0.5*(10 m/s/s)*(1 s)2 y = 20 m + (-5 m) y = 15 m | x = vix * t where vix = 34.6 m/s x = (34.6 m/s) * (1 s) x = 34.6 m |
|
The following table lists the results of such calculations for the first four seconds of the projectile's motion.
| Time | Horizontal Displacement | Vertical Displacement |
| 0 s | 0 m | 0 m |
| 1 s | 34.6 m | 15 m |
| 2 s | 69.2 m | 20 m |
| 3 s | 103.8 m | 15 m |
| 4 s | 138.4 m | 0 m |
The data in the table above show the symmetrical nature of a projectile's trajectory. The vertical displacement of a projectile t seconds before reaching the peak is the same as the vertical displacement of a projectile t seconds after reaching the peak. For example, the projectile reaches its peak at a time of 2 seconds; the vertical displacement is the same at 1 second (1 s before reaching the peak) as it is at 3 seconds (1 s after reaching its peak). Furthermore, the time to reach the peak (2 seconds) is the same as the time to fall from its peak (2 seconds).
Initial Velocity Components
It has already been stated and thoroughly discussed that the horizontal and vertical motions of a projectile are independent of each other. The horizontal velocity of a projectile does not effect how far (or how fast) a projectile falls vertically. Perpendicular components of motion are independent of each other. Thus, an
analysis of the motion of a projectile demands that the two components of motion are analyzed independent of each other, being careful not to mix horizontal motion information with vertical motion information. That is, if analyzing the motion to determine the vertical displacement, one would use kinematic equations with vertical motion parameters (initial vertical velocity, final vertical velocity, vertical acceleration) and not horizontal motion parameters (initial horizontal velocity, final horizontal velocity, horizontal acceleration). It is for this reason that one of the initial steps of a projectile motion problem is to determine the components of the initial velocity.
Earlier the method of vector resolution was discussed. Vector resolution is the method of taking a single vector at an angle and separating it into two perpendicular parts. The two parts of a vector are known as components and describe the influence of that vector in a single direction. If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and a vertical component. The horizontal velocity component (vx) describes the influence of the velocity in displacing the projectile horizontally; the vertical velocity component (vy) describes the influence of the velocity in displacing the projectile vertically. Thus, the analysis of projectile motion problems begins by using the trigonometric methods.
Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal. Such a projectile begins its motion with a horizontal velocity of 25 m/s and a vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity. These numerical values were determined by constructing a sketch of the vector with the given direction and using trigonometric functions to determine the sides of the triangle.
Determination of the Time of Flight
The time for a projectile to rise vertically to its peak (as well as the time to fall from the peak) is dependent upon vertical motion parameters. The process of rising vertically to the peak of a trajectory is a vertical motion and is thus dependent upon the initial vertical velocity and the vertical acceleration (g = 10 m/s/s, down). The process of determining the time to rise to the peak is an easy process - provided that you have a solid grasp of the concept of acceleration. When first introduced, it was said that acceleration is the rate at which the velocity of an object changes. An acceleration value indicates the amount of velocity change in a given interval of time. To say that a projectile has a vertical acceleration of -10 m/s/s is to say that the vertical velocity changes by 10 m/s (in the - or downward direction) each second. For example, if a projectile is moving upwards with a velocity of 40 m/s at 0 seconds, then its velocity will be 30 m/s after 1 second, 20 m/s after 2 seconds, 10 m/s after 3 seconds, and 0 m/s after 4 seconds. For such a projectile (with an initial vertical velocity of 40 m/s), it would take 4 seconds for it to reach the peak where its vertical velocity is 0 m/s. With this notion in mind, it is evident that the time for a projectile to rise to its peak is a matter of dividing the vertical component of the initial velocity by the acceleration of gravity.
Once the time to rise to the peak of the trajectory is known, the total time of flight can be determined. For a projectile which lands at the same height which it started, the total time of flight is twice the time to rise to the peak. Recall from the last section of Lesson 2 that the trajectory of a projectile is symmetrical about the peak. That is, if it takes 4 seconds to rise to the peak, then it will take 4 seconds to fall from the peak; the total time of flight is 8 seconds. The time of flight of a projectile is twice the time to rise to the peak.
Moderation Team
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