The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Projectile Motion Formulae Mathematics

In this explainer, we will learn how to derive formulae for projectile motion and use them in problems.

Suppose a particle is projected from a flat horizontal plane at an angle of πœƒβˆ˜ from the horizontal with an initial velocity of π‘ˆ mβ‹…sβˆ’1 and that no forces other than gravity act upon it during its flight.

Recall that we can decompose the particle’s velocity (or position or acceleration) into horizontal and vertical components by the formulas π‘ˆ=π‘ˆ(πœƒ),π‘ˆ=π‘ˆ(πœƒ)ο—ο˜cossin and that we can express these components in a velocity vector βƒ‘π‘ˆ=π‘ˆβƒ‘π‘–+π‘ˆβƒ‘π‘—ο—ο˜, where ⃑𝑖 and ⃑𝑗 are unit vectors in the horizontal and vertical directions.

Recall secondly that the particle obeys the following laws of motion.

Formula: Equations of Motion

If a particle is moving with initial velocity 𝑒 and constant acceleration π‘Ž, then its displacement 𝑠 at time 𝑑 is given by 𝑠=𝑒𝑑+12π‘Žπ‘‘.

Its velocity 𝑣 at time 𝑑 is given by 𝑣=𝑒+π‘Žπ‘‘.

The quantities displacement 𝑠 and velocity 𝑣 can be decomposed into horizontal and vertical components.

In the case of projectile motion, the initial velocity is π‘ˆ, there is no horizontal acceleration (i.e., π‘Ž=0), and the only vertical acceleration is due to gravity (i.e., π‘Ž=βˆ’π‘”ο˜). Therefore, 𝑠=π‘ˆπ‘‘+12π‘Žπ‘‘=π‘ˆ(πœƒ)𝑑,𝑠=π‘ˆπ‘‘+12π‘Žπ‘‘=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑,ο—ο—ο—οŠ¨ο˜ο˜ο˜οŠ¨οŠ¨cossin and 𝑣=π‘ˆπ‘₯+π‘Žπ‘‘=π‘ˆ(πœƒ),𝑣=π‘ˆ+π‘Žπ‘‘=π‘ˆ(πœƒ)βˆ’π‘”π‘‘.ο—ο—ο˜ο˜ο˜cossin

Consider the flight of the particle. It has a vertical displacement of zero at two times: its time of launch 𝑑=0 and its time of landing π‘‘οŠ§.

We can use the laws of motion to derive a formula for the particle’s flight time, π‘‘οŠ§. Consider the vertical component of its displacement, 𝑠=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑.sin

We want the time π‘‘οŠ§ when 𝑠=0, so 0=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑=π‘‘ο€»π‘ˆ(πœƒ)βˆ’π‘”2𝑑.sinsin

This equation has two solutions: the first solution 𝑑=𝑑=0 corresponds to the time of launch. The second solution is π‘ˆ(πœƒ)βˆ’π‘”2𝑑=0𝑔2𝑑=π‘ˆ(πœƒ)𝑑=2π‘ˆ(πœƒ)𝑔.sinsinsin

This is the formula for the flight time of the particle.

Formula: Total Flight Time

Suppose a particle is projected from a flat horizontal plane at an angle of πœƒβˆ˜ from the horizontal with an initial velocity of π‘ˆ mβ‹…sβˆ’1. Write 𝑔 for its vertical acceleration due to gravity and suppose that no forces other than gravity act upon it during its flight. Then, the particle’s time of landing π‘‘οŠ§, which is the same as its total flight time, is given by 𝑑=2π‘ˆ(πœƒ)𝑔.sin

The horizontal displacement of the particle at landing is called its range  𝑅.

We can substitute the formula for the total flight time into the expression for the horizontal displacement of the particle to derive a formula for 𝑅 as follows: 𝑅=𝑠=π‘ˆ(πœƒ)𝑑=π‘ˆ(πœƒ)Γ—2π‘ˆ(πœƒ)𝑔=2π‘ˆ(πœƒ)(πœƒ)𝑔.ο—οŠ§οŠ¨coscossinsincos

We can simplify this formula slightly using the double-angle identity 2(πœƒ)(πœƒ)≑(2πœƒ)sincossin to get 𝑅=π‘ˆ(2πœƒ)𝑔.sin

Formula: Range

Suppose a particle is projected from a flat horizontal plane at an angle of πœƒβˆ˜ from the horizontal with an initial velocity of π‘ˆ mβ‹…sβˆ’1. Write 𝑔 for its vertical acceleration due to gravity and suppose that no forces other than gravity act upon it during its flight. Then, the particle’s horizontal displacement at landing 𝑅, called its range, is given by 𝑅=π‘ˆ(2πœƒ)𝑔.sin

Suppose now we want to know the time the particle takes to reach its greatest height.

The particle is at its greatest height at the moment when it stops rising and starts falling, that is, when the vertical component of its velocity is zero. By taking 𝑣=0 in the formula for the vertical velocity, we have 0=π‘ˆ(πœƒ)βˆ’π‘”π‘‘π‘‘=π‘ˆ(πœƒ)𝑔.sinsin

This gives us the time taken to reach the greatest height. Observe that this time is half of the particle’s total flight time. This suggests that the particle’s trajectory is symmetrical about its highest point.

Formula: Greatest Height

Suppose a particle is projected from a flat horizontal plane at an angle of πœƒβˆ˜ from the horizontal with an initial velocity of π‘ˆ mβ‹…sβˆ’1. Write 𝑔 for its vertical acceleration due to gravity and suppose that no forces other than gravity act upon it during its flight. Then, the time 𝑑 at which the particle reaches its greatest height is given by 𝑑=π‘ˆ(πœƒ)𝑔.sin

Finally, let us derive an equation for the particle’s trajectory, that is, an equation giving the particle’s vertical position 𝑦 in terms of its horizontal position π‘₯. We have π‘₯=𝑠=π‘ˆ(πœƒ)𝑑cos and 𝑦=𝑠=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑.sin

We rearrange the first of these equations to make 𝑑 the subject as follows: π‘₯=π‘ˆ(πœƒ)𝑑𝑑=π‘₯π‘ˆ(πœƒ).coscos

Then, we substitute the expression π‘₯π‘ˆ(πœƒ)cos for 𝑑 in the second equation as follows: 𝑦=π‘ˆ(πœƒ)Γ—π‘₯π‘ˆ(πœƒ)βˆ’π‘”2Γ—ο€½π‘₯π‘ˆ(πœƒ).sincoscos

Now, we simplify, 𝑦=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆ(πœƒ),tancos and apply the Pythagorean identity sectan(πœƒ)≑1+(πœƒ): 𝑦=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆο€Ή1+(πœƒ).tantan

Observe that this equation is quadratic in π‘₯. This tells us that the flight path of a projectile moving freely under gravity is a parabola.

Formula: Position

Suppose a particle is projected from a flat horizontal plane at an angle of πœƒβˆ˜ from the horizontal with an initial velocity of π‘ˆ mβ‹…sβˆ’1. Write 𝑔 for its vertical acceleration due to gravity and suppose that no forces other than gravity act upon it during its flight. Write π‘₯ and 𝑦 for its horizontal and vertical position at time 𝑑. Then, the particle’s vertical position 𝑦 and horizontal position π‘₯ during flight are related by the quadratic equation 𝑦=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆο€Ή1+(πœƒ).tantan

While these formulae can be useful, it is not necessary to memorize them. It is much more important to understand how to derive them and related formulae by decomposing the projectile’s displacement, velocity, and acceleration into horizontal and vertical components and using the Newtonian laws of motion. For example, let us now derive a formula for the greatest height that the projectile reaches.

Example 1: Deriving Equations Related to the Greatest Height of a Projectile

A projectile is launched from a flat horizontal plane at an angle of πœƒ from the horizontal. Its initial velocity is π‘ˆ mβ‹…sβˆ’1, its acceleration due to gravity is 𝑔 mβ‹…sβˆ’2, and no other forces act upon it during its flight. Given that its greatest height is β„Ž=π‘Žπ‘ˆπ‘πœƒπ‘π‘”οŠ¨οŠ¨sinm, find the value of the constants π‘Ž, 𝑏, and 𝑐.

Answer

We consider the vertical component of the projectile’s displacement π‘ ο˜. We recall that the equation for vertical displacement is 𝑠=𝑒𝑑+12π‘Žπ‘‘, where π‘’ο˜ is the vertical component of the projectile’s initial velocity and π‘Žο˜ is the vertical component of its acceleration. We can express π‘’ο˜ in terms of the initial velocity π‘ˆ and the angle of projection πœƒ as follows: 𝑒=π‘ˆ(πœƒ).sin

We also know that the projectile has a vertical acceleration of βˆ’π‘” due to gravity. Therefore, 𝑠=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑.sin

If we can find the time 𝑑 at which the particle reaches its greatest height, then we can substitute into this equation to find the greatest height.

Recall that we can derive an expression for 𝑑 using the observation that this time is the time at which the vertical component π‘£ο˜ of the projectile’s velocity is zero. Using another equation of motion, 𝑣=𝑒+π‘Žπ‘‘ο˜ο˜, we get 0=π‘ˆ(πœƒ)βˆ’π‘”π‘‘π‘‘=π‘ˆ(πœƒ)𝑔.sinsin

We now substitute this time into 𝑠=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2π‘‘ο˜οŠ¨sin to get 𝑠=π‘ˆ(πœƒ)Γ—π‘ˆ(πœƒ)π‘”βˆ’π‘”2Γ—ο€½π‘ˆ(πœƒ)𝑔,sinsinsin and we simplify to get 𝑠=π‘ˆ(πœƒ)π‘”βˆ’π‘ˆ(πœƒ)2𝑔=π‘ˆ(πœƒ)2𝑔.sinsinsin

Therefore, π‘Ž=1, 𝑏=1, and 𝑐=2.

We can use projectile motion formulae to solve problems about projectiles.

Example 2: Finding the Starting Velocity of a Projectile given the Time of Flight and Starting Angle

A particle is projected from a horizontal plane at an angle of 47∘ from the horizontal. After 15 seconds, the particle hits the ground. Find the starting velocity of the particle, taking 𝑔=9.8/ms and giving your answer to 1 decimal place.

Answer

We can answer this question using the projectile motion formula for the total flight time, which is 𝑑=2π‘ˆ(πœƒ)𝑔.sin

Let us take a moment, however, to recall how that formula is derived.

We consider the vertical component of the projectile’s displacement π‘ ο˜. According to the equations of motion, 𝑠=𝑒𝑑+12π‘Žπ‘‘, where π‘’ο˜ is the vertical component of the projectile’s initial velocity and π‘Žο˜ is the vertical component of its acceleration. We can express π‘’ο˜ in terms of the angle of projection as follows: 𝑒=π‘ˆ(πœƒ).sin

We also know that the projectile has a vertical acceleration of βˆ’π‘” due to gravity. Therefore, 𝑠=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑=π‘‘ο€»π‘ˆ(πœƒ)βˆ’π‘”2𝑑.sinsin

We are interested in the time when the projectile hits the ground, so we set this vertical displacement to zero: π‘‘ο€»π‘ˆ(πœƒ)βˆ’π‘”2𝑑=0.sin

This equation has two solutions, the first solution, 𝑑=0, is the time of launch. The second solution corresponds to the projectile’s landing and gives the total flight time. We have 0=π‘ˆ(πœƒ)βˆ’π‘”2π‘‘π‘ˆ(πœƒ)=𝑔2𝑑,sinsin which gives us a relationship between the flight time and the initial velocity into which we can substitute the given values: π‘ˆ47=9.82Γ—15π‘ˆ=9.8Γ—15247=100.5β‹…sinsinms∘∘ to one decimal place.

In the following example, we use projectile motion formulae to determine what angle of projection yields the greatest range.

Example 3: Finding the Starting Angle that Results in the Greatest Range of a Projectile

A projectile is launched from a flat horizontal plane at an angle of πœƒ from the horizontal where 0β‰€πœƒβ‰€90∘∘. Its initial velocity is π‘ˆ m/s, its acceleration due to gravity is 𝑔 m/s2, and no other forces act upon it during its flight. Given that its range on the horizontal plane is 𝑅 m, find the value of πœƒ that maximizes 𝑅.

Answer

We are going to solve this question using the formula for a projectile’s range 𝑅=π‘ˆ(2πœƒ)𝑔.sin

Let us take a moment to recall how this formula is derived from the formula for a projectile’s total flight time.

We recall that the formula for the total flight time of a projectile is 𝑑=2π‘ˆ(πœƒ)π‘”οŠ§sin and that the horizontal component of its displacement can be expressed as 𝑠=π‘ˆ(πœƒ)𝑑.cos

Substituting the total flight time into this expression will give us the range 𝑅=π‘ˆ(πœƒ)Γ—2π‘ˆ(πœƒ)𝑔=2π‘ˆ(πœƒ)(πœƒ)𝑔.cossinsincos

Using the identity 2(πœƒ)(πœƒ)=(2πœƒ)sincossin, we have 𝑅=π‘ˆ(2πœƒ)𝑔.sin

This will reach its maximum when the function sin(2πœƒ) is at its maximum. We recall that the sine function has a maximum of 1 when its argument is 90∘.

Therefore, 𝑅 reaches its maximum value when 2πœƒ=90∘ or πœƒ=45∘.

If we know the coordinates of a point through which the projectile passes, then we can use the equation of its trajectory to calculate its angle of projection.

Example 4: Finding the Possible Starting Angles of a Projectile given the Initial Velocity and a Point it Passes Through

A particle is projected from a horizontal plane at an angle of πœƒ from the horizontal and with an initial velocity of 18 m/s. The particle passes through point 𝐴 situated 15 m horizontally and 8 m vertically from the point of launch. Find the possible values of πœƒ, taking 𝑔=9.8/ms and giving your answer to 1 decimal place.

Answer

We can solve this problem using the formula for a projectile’s trajectory 𝑦=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆο€Ή1+(πœƒ).tantan

Let us first take a moment to recall how this formula is derived. Recall that according to the laws of motion, the particle’s horizontal position π‘₯ and vertical position 𝑦 are given in terms of its initial velocity π‘ˆ, the angle with the horizontal πœƒ, and the time taken 𝑑 as follows: π‘₯=π‘ˆ(πœƒ)𝑑cos and 𝑦=π‘ˆ(πœƒ)π‘‘βˆ’π‘”2𝑑.sin

To derive the equation of trajectory, we want to write the equation for 𝑦 in terms of π‘₯ by using substitution. Rearranging the first equation, we have 𝑑=π‘₯π‘ˆ(πœƒ),cos which we can substitute into the second equation to get 𝑦=π‘ˆ(πœƒ)Γ—π‘₯π‘ˆ(πœƒ)βˆ’π‘”2Γ—ο€½π‘₯π‘ˆ(πœƒ)=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆ(πœƒ).sincoscostancos

Then, we apply the Pythagorean identity sectan(πœƒ)≑1+(πœƒ) for 𝑦=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆο€Ή1+(πœƒ).tantan

Notice that this is a quadratic equation in π‘₯, but it is also a quadratic equation in tan(πœƒ). We are going to solve for tan(πœƒ) using the coordinates (15,8) of the point through which the particle passes to get 8=15(πœƒ)βˆ’9.8Γ—152Γ—18ο€Ή1+(πœƒ).tantan

Expanding and simplifying, we have 3.402Μ‡7ο€Ή1+(πœƒ)ο…βˆ’15(πœƒ)+8=03.402Μ‡7(πœƒ)βˆ’15(πœƒ)+11.402Μ‡7=0,tantantantan whose solutions we can find using the quadratic formula βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž=15±√15βˆ’4Γ—3.402Μ‡7Γ—11.402Μ‡72Γ—3.402Μ‡7=15±√225βˆ’155.204486.80Μ‡5, giving tan(πœƒ)=3.4317 and tan(πœƒ)=0.9765. Applying tan, we have πœƒ=73.8∘ or πœƒ=44.3∘ to one decimal place.

We will now apply the formulae of projectile motion to a real-world problem.

Example 5: Solving Real-World Problems with Projectile Motion Formulae

A rocket is launched vertically at a speed of 60 m/s from a point 𝑋. When it reaches its maximum height, a capsule is ejected horizontally from it at a speed of 40 m/s. Find the horizontal distance from 𝑋 to the capsule’s landing point, taking 𝑔=9.8/ms and giving your answer to 1 decimal place.

Answer

We decompose the motion of the capsule into its horizontal and vertical components. Since the capsule is launched horizontally when the rocket is at its maximum height, its vertical velocity at this point is 0 and it has a constant acceleration of 𝑔=9.8/ms in the negative 𝑦-direction. The rocket also has vertical velocity 0 and constant acceleration 𝑔=9.8/ms at this point. Therefore, both rocket and capsule will take the same amount of time to reach the ground.

In this model, the only force acting on the rocket is that of gravity. The rocket reaches its maximum height when its deceleration due to gravity reduces its initial vertical velocity to zero. Since gravity is the only force in play, this takes the same amount of time as it takes for the rocket to accelerate from zero at its highest point back to its initial velocity downward. At this point, it hits the ground.

Thus, the capsule will take the same amount of time to reach the ground as the rocket, which is equal to the amount of time the rocket took to reach its maximum height.

Recall that the time taken by the rocket to reach its maximum height is given by the formula 𝑑=π‘ˆ(πœƒ)𝑔,sin where π‘ˆ is the initial speed and πœƒ is the angle of projection. We have that π‘ˆ=60 and the rocket is launched vertically, so πœƒ=90∘ and sin(πœƒ)=1. Therefore, 𝑑=609.8=6.122.s

We can substitute this time into the equation of motion governing the horizontal displacement of the capsule 𝑠=π‘ˆπ‘‘, where 𝑒 is the horizontal component of its initial velocity. We have that 𝑒=40. Therefore, 𝑠=40Γ—6.122=244.9m to one decimal place.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We can decompose the motion of a projectile into horizontal and vertical components and apply the equations of motion 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨ and 𝑣=𝑒+π‘Žπ‘‘ to derive formulae for the motion of a projectile.
  • The total flight time π‘‘οŠ§ of a projectile is given by 𝑑=2π‘ˆ(πœƒ)𝑔,sin where π‘ˆ is its initial velocity, πœƒ is its angle of projection, and 𝑔 is the acceleration due to gravity.
  • The projectile’s range 𝑅 is given by 𝑅=π‘ˆ(2πœƒ)𝑔.sin
  • The time 𝑑 at which the projectile reaches its greatest height is given by 𝑑=π‘ˆ(πœƒ)𝑔.sin
  • The projectile’s vertical position 𝑦 and horizontal position π‘₯ during flight are related by the quadratic equation 𝑦=π‘₯(πœƒ)βˆ’π‘”π‘₯2π‘ˆο€Ή1+(πœƒ).tantan
  • We can use projectile motion formulae to answer questions about the trajectories of projectiles.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.