Lesson Explainer: Projectile Motion Physics • 9th Grade

In this explainer, we will learn how to analyze the motion of objects that move horizontally while undergoing constant vertical acceleration.

Recall that projectiles are objects that have nonzero uniform vertical acceleration while moving horizontally at constant velocity. Projectile motion refers to the motion of any projectileβ€”for example, a rock that is dropped from a height or a ball that is thrown.

If the projectile is traveling at high speed, then we may have to consider the effects of air resistance, but most projectiles that we will deal with travel slow enough for us to ignore these effects. This means that the only force acting on the object is gravity, which acts downward. The object’s horizontal velocity is therefore constant while it is in motion.

The diagram below shows a projectile of mass π‘š launched with a launch speed of 𝑣 at an angle πœƒ from the horizontal.

The horizontal velocity of the projectile, 𝑣, is equal to π‘£πœƒcos. Similarly, the vertical velocity of the projectile, π‘£ο˜, is equal to π‘£πœƒsin.

Three important quantities that we would like to calculate are

  • time of flight,
  • range (horizontal distance traveled by the projectile),
  • maximum altitude (the maximum vertical distance traveled by the projectile).

These three quantities are illustrated for a projectile being launched from the same vertical displacement at which it finishes the motion in the diagram below.

The time of flight of the projectile can be found by starting with the equation for displacement, 𝑠, with constant acceleration, π‘Ž, and initial velocity, 𝑒, over time, 𝑑: 𝑠=𝑒𝑑+12π‘Žπ‘‘.

We will consider the vertical motion only, so we can substitute our initial velocity with initial vertical velocity, 𝑒=π‘£ο˜, and the constant acceleration with gravity, π‘Ž=βˆ’π‘”. Note that gravity acts downward, which is negative in the vertical direction, so it is a negative term in our calculations. This gives us an equation for vertical displacement, π‘ ο˜: 𝑠=π‘£π‘‘βˆ’12𝑔𝑑.

We can take 𝑑 as a common factor: 𝑠=π‘‘ο€Όπ‘£βˆ’12π‘”π‘‘οˆ.

We would like to know when the vertical displacement of the projectile is zero, 𝑠=0: 0=π‘‘ο€Όπ‘£βˆ’12π‘”π‘‘οˆ.

This tells us that there are two times when the vertical displacement is zero: one at the beginning of the projectile motion, and one at the end of the projectile motion.

At the end of the motion, time 𝑑=𝑇. The second solution can be found when π‘£βˆ’12𝑔𝑇=0.

Then, we rearrange to make 𝑇 the subject of the equation: 𝑇=2𝑣𝑔.

Finally, inserting the initial vertical velocity, 𝑣=π‘£πœƒο˜sin, gives the equation for time of flight: 𝑇=2𝑣(πœƒ)𝑔.sin

Definition: Time of Flight

When the final vertical displacement of the projectile is equal to the initial vertical displacement, the time of flight, 𝑇, can be calculated as 𝑇=2𝑣(πœƒ)𝑔,sin where 𝑣 is the initial speed of the projectile, πœƒ is the launch angle measured above the horizontal, and 𝑔 is the gravitational constant.

Let’s work through an example question where we are required to calculate the time of flight of a projectile.

Example 1: Calculating the Time of Flight of a Projectile

A projectile has an initial speed of 25 m/s and is fired at an angle of 48∘ above the horizontal. What is the time between the projectile leaving the ground and returning to the ground at the same height that it was launched from?

Answer

Let’s start by drawing a diagram of this scenario.

The question is asking for the time of flight of the projectile.

To calculate the time of flight, we use the following formula: 𝑇=2𝑣𝑔=2𝑣(πœƒ)𝑔.sin

As stated in the question, the projectile is fired at an angle of πœƒ=48∘ above the horizontal with an initial speed of 𝑣=/ms, so the vertical velocity of the projectile is calculated as 𝑣=𝑣(πœƒ)=25(48)sinsin resulting in a vertical velocity (to two decimal places) of 𝑣=18.58/.ms

Substituting this into our time of flight equation, with the gravitational constant 𝑔=9.8/ms, gives us 𝑇 (to two decimal places): 𝑇=2Γ—18.589.8=3.79/.ms

We can use the equation for the time of flight of the projectile to calculate the horizontal distance traveled, also known as the range. Because there is no force acting on the object in the horizontal direction, the equation for horizontal displacement, 𝑠=𝑅, of the projectile at the end of the motion is simply the horizontal velocity of the projectile, 𝑣, multiplied by the time of flight, 𝑇: 𝑅=𝑣𝑇.

Inserting horizontal velocity, 𝑣=π‘£πœƒο—cos, and the equation for time of flight, 𝑇=2π‘£πœƒπ‘”sin, we get 𝑅=(π‘£πœƒ)ο€½2π‘£πœƒπ‘”ο‰.cossin

This gives us our final equation for the range of the projectile: 𝑅=2𝑣(πœƒ)(πœƒ)𝑔.sincos

Using the formula sinsincos(2πœƒ)=2(πœƒ)(πœƒ), the range of the projectile can also be written as 𝑅=𝑣(2πœƒ)𝑔.sin

Definition: Horizontal Range

The horizontal range, 𝑅, of a projectile launched from the same initial and final vertical displacement can be calculated as 𝑅=2𝑣(πœƒ)(πœƒ)𝑔,sincos where 𝑣 is the initial speed of the projectile, πœƒ is the launch angle measured above the horizontal, and 𝑔 is the gravitational constant.

Let’s look at an example question where we are required to calculate the range of a projectile.

Example 2: Calculating the Range of a Projectile

A projectile has an initial speed of 15 m/s at a launch angle of 28∘ above the horizontal. What is the horizontal displacement of the projectile from its launch position to where it lands if its vertical displacement from its launch position is zero?

Answer

This question requires us to calculate the range of the projectileβ€”the horizontal displacement of the projectile when it returns to a vertical displacement of zero from its launch position.

We can start with our equation for range: 𝑅=2𝑣(πœƒ)(πœƒ)𝑔.sincos

We can immediately substitute in the values given to us for 𝑣=15/ms, πœƒ=28∘, and 𝑔=9.8/ms: 𝑅=2Γ—15Γ—(28)Γ—(28)9.8.sincos

This gives us our horizontal displacement of 𝑅=19.03.m

We can calculate the maximum altitude of the projectile by considering the peak of the trajectory. This occurs when the vertical velocity of the projectile is zero.

Here we can start with another constant acceleration equation that relates velocity during the motion, 𝑣, to the initial velocity, 𝑒; the constant acceleration, π‘Ž; and the displacement of the object from the start of the motion, 𝑠: 𝑣=𝑒+2π‘Žπ‘ .

When calculating maximum altitude, we would like to find the vertical displacement, 𝑠=π‘ ο˜, of the projectile given an initial vertical velocity of 𝑒=π‘£ο˜ at the point when the vertical velocity of the projectile is equal to zero, 𝑣=0. At this point, the vertical displacement of the projectile is at its maximum altitude, which we will write as 𝑠=β„Žο˜. As in previous calculations, gravitational acceleration acts in the negative vertical direction, so π‘Ž=βˆ’π‘”: 0=π‘£βˆ’2π‘”β„Ž.

We can rearrange this equation to make π‘ ο˜ its subject by first adding 2π‘”β„Ž to both sides: 2π‘”β„Ž=𝑣.

Then, dividing both sides by 2𝑔, we get β„Ž=𝑣𝑔.

We can then insert our expression for the initial vertical velocity of a projectile launched at speed 𝑣 at an angle πœƒ above the horizontal, 𝑣=π‘£πœƒο˜sin. This gives us our final equation for maximum altitude: β„Ž=𝑣(πœƒ)𝑔.sin

Definition: Maximum Altitude

The maximum altitude, β„Ž, of a projectile can be calculated as β„Ž=𝑣(πœƒ)𝑔,sin where 𝑣 is the initial speed of the projectile, πœƒ is the launch angle measured above the horizontal, and 𝑔 is the gravitational constant.

Now let’s look at an example where we must work backward from maximum altitude to work out the angle above the horizontal that the projectile was launched at.

Example 3: Calculating the Launch Angle of a Projectile from Maximum Altitude

A projectile is fired with an initial speed of 28 m/s and has a maximum upward vertical displacement from its launch position of 4.4 m. What angle above the horizontal is the projectile launched at?

Answer

This question requires us to work backward from maximum altitude to find the angle above the horizontal the projectile is launched at.

We can start with our equation for maximum altitude: β„Ž=𝑣(πœƒ)𝑔.sin

We must rearrange this to get an equation for the angle above the horizontal, πœƒ. Let’s us start by multiplying both sides of the equation by 2𝑔: 2π‘”β„Ž=𝑣(πœƒ).sin

Then, we can take the square root of both sides: √2π‘”β„Ž=π‘£πœƒ.sin

Next, we can divide both sides by 𝑣: sinπœƒ=√2π‘”β„Žπ‘£.

Finally, we take the inverse sine of both sides πœƒ=ο€Ώβˆš2𝑔𝑠𝑣.sin

Now we can substitute in our values for maximum altitude, 𝑠=4.4m, and initial speed, 𝑣=28/ms, and the gravitational constant, 𝑔=9.8/ms: πœƒ=ο€Ώβˆš2Γ—9.8Γ—4.428.sin

This gives us our launch angle above the horizontal: πœƒ=19.3.∘

We should also be comfortable with the changes in mechanical energy during projectile motion.

With no air resistance, mechanical energy is conserved throughout the motion. That means that the sum of the kinetic energy and the gravitational potential energy at any point during the motion is constant: ()+()=()+().KEPEKEPE

However, when air resistance is significantβ€”when the projectile is traveling at high speedsβ€”some mechanical energy is lost during the motion.

The force from air resistance changes with the speed at which the projectile is traveling, which is nonlinear so the mechanical energy loss throughout the motion is also nonlinear.

Let’s now look at a question about the mechanical energy of a projectile.

Example 4: The Mechanical Energy of a Projectile during Motion

The change in the vertical displacement of a projectile from its launch position and its horizontal displacement from its launch position is shown in the graph. Which of the following graphs most correctly represents how the mechanical energy of the projectile changes from the instant that it is launched to the instant that it lands?

Answer

When we look at the graph of vertical and horizontal displacement, we can immediately see that the projectile has slowed down during the motion. This means that it has lost some mechanical energy to air resistance. This rules out graph V.

Next, we can see that the projectile is still in motion when it completes its trajectory; this means that it still has some mechanical energy at the end of the motion. This means that graphs II and IV cannot correctly represent the mechanical energy of the projectile.

This leaves us with graphs I and III: both show the mechanical energy decreasing over time, still greater than zero when the projectile lands. However, graph I is nonlinear, and graph III is linear.

Recall that air resistance is highly nonlinear, causing nonlinear mechanical energy loss. This means that graph III cannot correctly represent the mechanical energy of the projectile.

This means that graph I correctly represents the mechanical energy of the projectile from the instant that it is launched to the instant that it lands.

When the launch point of the projectile is above or below its landing point, we must consider the general equation for vertical displacement of the projectile: 𝑠=π‘£π‘‘βˆ’12𝑔𝑑.

As before, π‘ ο˜ is the vertical displacement of the projectile, π‘£ο˜ is the initial vertical velocity of the projectile, 𝑑 is time, and 𝑔 is the gravitational constant.

The vertical displacement of the landing point is different from that of the launch point. We will write this difference as 𝑠=π‘‘ο˜. We can substitute this into the equation for vertical displacement at time 𝑑=𝑇, the time of flight of the projectile: 𝑑=π‘£π‘‡βˆ’12𝑔𝑇.

Note that 𝑑 will be positive if the landing point of the projectile is above its launch point and negative if the landing point is below the launch point.

To find the time of flight, we can first rearrange the equation to be in standard quadratic form in terms of 𝑇: 12π‘”π‘‡βˆ’π‘£π‘‡+𝑑=0.

We can solve this quadratic equation to find the time of flight, 𝑇.

The horizontal range of the projectile, 𝑅, is then calculated as before: 𝑅=𝑣𝑇, where 𝑣 is the initial horizontal velocity of the projectile and 𝑇 is the time of flight.

Finally, let’s work through a slightly more complex question, where the launch point of the projectile is above its landing point.

Example 5: Calculating the Range of a Projectile Launched from an Elevated Point

A rock is thrown horizontally from a height of 9.6 m from the ground with a speed of 5.2 m/s. Calculate the horizontal range of the rock. Use 𝑔=10/ms.

  1. 9.6 m
  2. 2.4 m
  3. 7.2 m
  4. 16.8 m

Answer

In this question, we must calculate the horizontal range of a projectile launched from an elevated point.

First, we must calculate the time of flight of the projectile. We will start with our initial equation for vertical displacement: 𝑠=π‘£π‘‘βˆ’12𝑔𝑑.

Knowing that the final vertical displacement is 9.6 m Β lower than the launch point and the initial vertical velocity is zero, we can substitute in values for 𝑠=βˆ’9.6, 𝑣=0, and 𝑔=10. This is reached at the end of the motion, so time 𝑑=𝑇: βˆ’9.6=βˆ’12Γ—10×𝑇.

First, we rearrange for π‘‡οŠ¨: 𝑇=9.6Γ—210=1.92.

Then, we take the square root to get the time of flight of the projectile (to two decimal places): 𝑇=√1.92=1.38.s

Now we can calculate the range of the projectile, knowing that its horizontal velocity remains constant throughout the motion.

The range of the projectile is equal to 𝑅=𝑣𝑇.

In this case, the projectile was launched horizontally, so 𝑣=5.2/ms. Substituting in our values for 𝑣 and 𝑇, 𝑅=5.2Γ—1.38=7.21.m

So, the correct answer is C 7.2 m (to one decimal place).

Key Points

  • Generally, the only force acting on a projectile is gravity:
    • This means vertical velocity experiences a constant acceleration of 𝑔 vertically downward.
    • Horizontal velocity remains constant throughout the motion.
  • The vertical displacement of a projectile can be found using the equation for displacement with a constant acceleration of βˆ’π‘”:
    • 𝑠=π‘£π‘‘βˆ’12π‘”π‘‘ο˜ο˜οŠ¨.
  • The horizontal displacement of a projectile can be found using the equation for displacement with a constant acceleration of zero:
    • 𝑠=𝑣𝑑.
  • The time of flight is the total duration of the projectile’s motion.
  • The range is the horizontal displacement of the projectile at the end of its motion.
  • The maximum altitude is the maximum vertical displacement of the projectile.
  • Mechanical energy throughout projectile motion is conserved (remains constant) unless air resistance is not negligible, where mechanical energy is lost nonlinearly over time.

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