Lesson Explainer: Projectiles Physics • 9th Grade

In this explainer, we will learn how to analyze the motion of projectiles: objects that have nonzero uniform vertical acceleration while moving horizontally at constant velocity.

Understanding the motion of projectiles depends on understanding that the vertical motion of a projectile is different to its horizontal motion.

The horizontal motion of a projectile is uniform. Consider an object that is moving uniformly horizontally and has zero vertical velocity. If the position of the object is recorded at equal time intervals, the positions would be equally separated, as is the case in the following figure.

The vertical motion of a projectile is uniformly accelerated. Consider an object that is accelerating uniformly vertically downward and has zero horizontal velocity. An example of this would be a ball dropped from the top of a building. If the position of the object is recorded at equal time intervals, the positions would be separated by distances that increase with successive time intervals, as is the case in the following figure.

Suppose now that for each horizontal position, all of the vertical positions are shown, and the positions in both horizontal and vertical directions at successive instances are highlighted, as shown in the following figure.

Finally, let us indicate the path that the object would follow between the positions shown, as the blue curve in the following figure shows.

What we can see in the figure is the trajectory of a projectile that has no initial vertical velocity. An example of this kind of trajectory would be that of a stone thrown horizontally from the top of a cliff, as shown in the following figure.

Let us now look at an example in which the vertical and horizontal motion of a projectile is considered.

Example 1: Comparing the Horizontal and Vertical Motion of a Projectile

The diagram shows the changes in the horizontal and vertical displacement of an object at equal time intervals. Gravity is the only force acting on the object.

  1. Is the object’s horizontal speed increasing, decreasing, or constant?
  2. Is the object’s vertical speed increasing, decreasing, or constant?
  3. Is the object’s horizontal acceleration increasing, decreasing, or constant?
  4. Is the object’s vertical acceleration increasing, decreasing, or constant?
  5. Is the object’s total speed increasing, decreasing, or constant?

Answer

Part 1

The horizontal speed of the object is the rate of change of horizontal distance moved by the object with time. The diagram shows that the horizontal distance moved by the object is the same in each time interval, hence the horizontal speed is constant.

The answer to part one is “constant.”

Part 2

The vertical speed of the object is the rate of change of vertical distance moved by the object with time. The diagram shows that the vertical distance moved by the object increases in successive time intervals, so the vertical speed of the object is increasing.

The answer to part two is “increasing.”

Part 3

The horizontal acceleration of the object is the rate of change of horizontal velocity with time.

We have seen from the first part of the question that the horizontal velocity is constant, and for this reason the horizontal acceleration must be zero throughout the motion of the object. As the horizontal acceleration is zero throughout the motion, the value of the horizontal acceleration is constant.

The answer to part three is “constant.”

Part 4

The vertical acceleration of the object is the rate of change of vertical velocity with time.

We have seen from the second part of the question that the vertical velocity increases in successive time intervals. We cannot tell clearly from the diagram whether vertical velocity increases in successive time intervals as the diagram shows the vertical displacement rather than the vertical velocity.

The question states however that the only force acting on the object is gravity. Gravitational force from Earth on an object is approximately constant near the surface of Earth, so the value of the vertical acceleration is constant.

The answer to part four is “constant.”

Part 5

The object is increasing in speed in the vertical direction. For the object to have the same total speed, where this is the speed of the object in the direction of its motion, the speed of the object in the horizontal direction would have to be decreasing proportionally to the increase in the speed in the vertical direction. We have seen that the horizontal speed is constant, so we see that the speed of the object in its direction of motion must be increasing.

The answer to part five is “increasing.”

The difference between the vertical and horizontal speeds of a projectile throughout its motion can be analyzed using kinematic graphs.

Let us look at an example where the vertical and horizontal speeds of a projectile throughout its motion are analyzed using kinematic graphs.

Example 2: Identifying Speed–Time Graphs for a Projectile

An object is given a short horizontal push that sets it in motion along a smooth horizontal surface. When the object reaches the end of the surface, it undergoes projectile motion from an initial position to a final position, as shown in the diagram.

  1. Which of the graphs (a), (b), (c), and (d) shows the changes in the vertical speed of the object between its initial and final positions?
  2. Which of the graphs (e), (f), (g), and (h) shows the changes in the horizontal speed of the object between its initial and final positions?

Answer

Part 1

The object undergoes projectile motion, so it accelerates downward uniformly throughout its motion. The object initially moves only horizontally due to the short horizontal push given to it. The initial vertical speed of the object is therefore zero. From this we see that the correct graph must be either (b) or (d), as (a) and (c) both show the initial vertical speed as nonzero.

The object accelerates downward uniformly throughout its motion. Uniform acceleration results in equal changes in speed in equal time intervals. The slope of a speed–time graph has a slope 𝑔, where 𝑔 is given by 𝑔=Δ𝑣Δ𝑡, where 𝑣 is the speed and 𝑡 is the time.

A speed–time graph showing equal changes in speed in equal time intervals must show a line of constant slope, which is a straight line. Only graph (d) has a straight line, so is the correct graph.

The answer to part one is graph (d).

Part 2

The acceleration of the object throughout its motion is entirely vertical. The initial horizontal velocity of the object does not increase throughout the motion. This immediately rules out graph (h). The question does not indicate the presence of any force that acts horizontally on the object other than the force responsible for the initial horizontal velocity of the object. The two graphs that show the final horizontal speed of the graph as zero cannot then be correct, so graph (e), showing constant horizontal speed, is correct.

The answer to part two is graph (e).

We have so far only considered projectiles that have no initial vertical velocity. To have the trajectory of a projectile, such an object must have at least an initial vertical displacement. Let us now consider a projectile that has no initial vertical displacement and which must have an initial vertical velocity.

A projectile that has an initial upward velocity is uniformly accelerated in the opposite direction to its initial motion. The upward velocity decreases until it reaches zero, at which point the projectile is instantaneously at rest. After this instant, the projectile moves as we have already seen a projectile with zero initial vertical velocity move.

The vertical acceleration and velocity along the trajectory of a projectile with an initial upward velocity is shown in the following figure.

The part of the trajectory for which the projectile is ascending and the part of the trajectory for which the projectile is descending are symmetrical. The vertical speed of a projectile is the same at a given height whether the projectile is ascending or descending.

A projectile that is just released to fall downward is acted on only by gravity. A projectile that moves horizontally or vertically upward, though, must have a force other than gravity act on it as it is launched. However, this force does not act on the projectile once it has been launched. For both projectiles that are released and those that are launched, a projectile is modeled as starting to move with an initial velocity that thereafter only changes due to gravitational acceleration.

The force launching a projectile has a vertical and a horizontal component. The greater the component of the force in a direction, the greater the initial velocity of the projectile in that direction. The magnitudes of these velocities determine the trajectory of a projectile.

Let us now look at an example comparing projectiles that follow different trajectories.

Example 3: Comparing the Motion of a Projectile Following Different Trajectories

The object shown in the diagram is undergoing two examples of projectile motion due to the forces 𝐹 and 𝐹 and due to the object’s weight. The first example of projectile motion, example A, involves a launch angle for the object that is close to being vertically upward. The other example, example B, shows a launch angle that is much shallower, closer to being horizontal. For both examples, the horizontal distance traveled by the object is the same.

  1. Which example of projectile motion involves a greater vertical velocity?
  2. Which example of projectile motion involves the object being airborne for a longer time?
  3. Which example of projectile motion involves a greater horizontal velocity?

Answer

Part 1

In example A, the projectile travels a greater upward vertical distance. In both examples, the projectiles have the same constant downward acceleration due to gravity, and so the upward vertical velocity must be greater in example A.

The answer to part one is example A.

Part 2

The time for which a projectile is airborne is the time taken for the upward vertical motion of the projectile to cease and the projectile to then fall back to the ground. When the projectiles are at their highest points, the projectile in example A has a greater vertical distance to travel to reach the ground than the projectile in example B. We see then that the projectile in example A must take more time to fall back to the ground from its highest point than the projectile in example B.

This shows that the projectile in example A is airborne for a greater time.

The answer to part two is example A.

Part 3

The horizontal distance from the launch point to the landing point is the same in both examples. This might make it seem that the horizontal velocity of the projectile is the same in both examples.

However, this is incorrect. This can be seen by recalling that the time for which the object moves is not equal in the examples and by understanding that this difference in times is related to the maximum height reached by the object in each example.

The midpoint of the trajectory of a projectile is where the projectile is at its greatest height and where it has zero vertical speed, as shown in the following figure.

In the question, we see that the projectile is launched from and lands at the same point in both examples. This means that the midpoint of the trajectory of the projectile is halfway from the launching point and the landing point in both examples.

Knowing that the projectile in example A is airborne for more time than the projectile in example B and also knowing that the projectile in example A travels the same horizontal distance as the projectile in example B, we can compare the horizontal speeds of the projectiles.

Distance, time, and speed are related by the formula distanceSpeedtime=×.

In example A, we can see that 𝑑=𝑣𝑡, and in example B, we can see that 𝑑=𝑣𝑡, where 𝑣 represents the horizontal speeds of the projectiles and 𝑡 represents the times they were airborne in examples A and B and 𝑑 represents the horizontal distance they moved.

The horizontal distance moved by the projectiles is the same, so we can say that 𝑣𝑡=𝑣𝑡.

We can divide this equation by 𝑡. Doing this, we see that 𝑣=𝑣𝑡𝑡.

We have shown that 𝑡>𝑡.

It must then be the case that 𝑡𝑡<1.

From this it follows that 𝑣<𝑣.

The horizontal velocity is greater for example B.

Let us now look at an example involving analyzing just the horizontal motion of a projectile launched upward.

Example 4: Identifying Velocity–Time Graphs for a Projectile

An object is set in motion by an initial force 𝐹 that acts diagonally upward, as shown in the diagram. The object undergoes projectile motion.

  1. Which of the graphs (a), (b), (c), and (d) shows the changes in the horizontal displacement of the object between leaving the ground and returning to the ground?
  2. Which of the graphs (e), (f), (g), and (h) shows the changes in the horizontal velocity of the object between leaving the ground and returning to the ground?

Answer

Part 1

Initially, the projectile has zero horizontal displacement. Graph (d) shows nonzero horizontal displacement, so it can be immediately ruled out. If graph (d) had shown an initial zero displacement, it might seem that it could have been correct, but this is not the case, as graph (d) shows a constant value for displacement, meaning that the projectile did not move horizontally. This is only correct for a projectile launched completely vertically, and the diagram in the question shows that the projectile is launched by a force that does not act completely vertically.

The horizontal component of the force that launches the projectile is responsible for the initial horizontal velocity of the projectile. After the projectile is launched, its horizontal velocity is uniform.

The slope of a displacement–time graph has a slope 𝑣, where 𝑣 is given by 𝑣=Δ𝑠Δ𝑡, where 𝑠 is the displacement and 𝑡 is the time. The slope 𝑣 equals the velocity of the object plotted by the graph. Graphs (a) and (b) both show changing slopes, so they cannot represent the horizontal velocity of the projectile. Graph (c) shows a constant slope, so it can be a correct representation of how the horizontal displacement of the projectile changes.

The answer to part one is graph (c).

Part 2

We have established that the horizontal velocity of the projectile is constant, so a velocity–time graph for the projectile’s horizontal motion should have a constant value. Only graph (e) has a constant value. Graph (g) has a constant slope, but this would show a constantly increasing velocity rather than a constant velocity.

The answer to part two is graph (e).

Let us now look at an example involving analyzing just the vertical motion of a projectile launched upward.

Example 5: Identifying Displacement–Time and Velocity–Time Graphs for a Projectile

An object is set in motion by an initial force 𝐹 that acts diagonally upward, as shown in the diagram. The object undergoes projectile motion.

  1. Which of the graphs (a), (b), (c), and (d) shows the changes in the vertical displacement of the object between leaving the ground and returning to the ground? Consider vertically upward displacement as positive.
  2. Which of the graphs (e), (f), (g), and (h) shows the changes in the vertical velocity of the object between leaving the ground and returning to the ground? Consider vertically upward displacement as positive.

Answer

Part 1

Initially, the projectile has zero vertical displacement. All the graphs show this, so none can be ruled out immediately. the projectile eventually lands, and when it lands, its vertical displacement is equal to its vertical displacement when it launches. Only graphs (c) and (d) correctly show this.

A graph of the vertical displacement of a projectile against time will have approximately the same shape as the trajectory of a projectile, the only difference being that on such a graph, the 𝑥-axis of the graph will be scaled according to the time for which the projectile has a given vertical displacement, whereas a trajectory shows what horizontal displacement the projectile has for a given vertical displacement. Graph (c) resembles a projectile’s trajectory, whereas graph (d) shows the projectile initially accelerating upward before decelerating. The acceleration of the projectile is downward throughout its motion, so graph (d) is incorrect. Graph (c) is correct.

The answer to part one is graph (c).

Part 2

The initial vertical velocity of the projectile is nonzero. The projectile is moving upward when it launches. Graphs (e) and (f) can therefore be immediately ruled out.

Graphs (g) and (h) both correctly show that the vertical velocity of the projectile is zero at the instant midway between when it launches and when it lands. Graph (g) shows that the speed of the projectile is the same when it lands as when it launches, which is correct, but the velocity of the projectile is different when it launches and when it lands as the directions of the launching and landing velocity are opposite. Graph (h) shows that the initial velocity is positive, which the question states upward velocity should be taken as, and finally negative, showing a reversal of the vertical direction of the motion of the projectile. The vertical acceleration of the projectile is constant and downward; hence, the slope of the velocity–time graph should be a downward-sloping straight line, which is true of the slope of (h). We see then that (h) is correct.

The answer to part two is graph (h).

Let us now summarize what we have learned in these examples.

Key Points

  • A projectile has a constant horizontal velocity and a constant vertical acceleration, assuming that no air resistance acts on it.
  • The trajectory of a projectile is a curved path.
  • The time for which a projectile is airborne depends on its initial vertical velocity.
  • The horizontal distance traveled by a projectile depends on the horizontal velocity of the projectile and the time for which it remains airborne.
  • The vertical distance traveled by a projectile depends on the initial vertical velocity of the projectile.
  • A projectile has zero vertical velocity when at its greatest vertical displacement.
  • A projectile that launches and lands at points of the same altitude has the same speed at a given height above this altitude whether the projectile is ascending or descending.

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