How to Solve Projectile Motion Problems: Applying Newton's Equations of Motion to Ballistics

© Eugene Brennan

Physics is an area of science which deals with how matter and waves behave in the universe. A branch of physics called mechanics deals with forces, matter, energy, work done and motion. A further sub-branch known as kinematics deals with motion and ballistics is specifically concerned with the motion of projectiles launched into the air, water or space.

Solving ballistic problems involves using the kinematics equations of motion, also known as the SUVAT equations or Newton's equations of motion.

In these examples, for the sake of simplicity, the effects of air friction (known as drag) have been excluded.

See also my article, Newton's 3 Laws of Motion: Force, Mass and Acceleration, which introduces the basic concepts.

What Are the Equations of Motion? (SUVAT Equations)

Consider a body of mass m, acted on by a force F for time t. This produces an acceleration which we will designate with the letter a. The acceleration increases the velocity of the body from its initial value of u, and after time t, it reaches a velocity v. It also travels a distance s.
So we have five parameters associated with the body in motion: s, u, v, a and t.
If the force F is constant, the acceleration a is constant and velocity increases linearly over time t.

Acceleration of body. A force F produces an acceleration a over time t and distance s. The acceleration increases the velocity from u to v. © Eugene Brennan

The equations of motion allow us to work out any of these parameters once we know three other parameters. So the three most useful formulae are:

Equations of motion

Remember, Newton's second law of motion tells us that F = ma, so the acceleration of a body depends on the force applied F and its mass m. The body only accelerates and its velocity increases as long as a force is applied (or decelerates and decreases in velocity if the force opposes motion). Once the force is removed, the velocity of the body stays constant unless another force acts on it (Newton's first law of motion). In our examples, that other force is gravity, which causes velocity to increase or decrease.

Solving Projectile Motion Problems: Calculating Time of Flight, Distance Travelled and Altitude

High school and college exam questions in ballistics usually involve calculating time of flight, distance travelled and altitude attained.

There are four basic scenarios normally presented in these types of problems, and it is necessary to calculate the parameters mentioned above:

•  Object dropped from a known altitude
•  Object thrown upward
•  Object thrown horizontally from a height above the ground
•  Object launched from the ground at an angle

These problems are solved by considering the initial or final conditions and this enables us to work out a formula for velocity, distance travelled, time of flight and altitude. To decide which of Newton's three equations to use, check which parameters you know and use the equation with one unknown – i.e., the parameter you want to work out.

In examples three and four, breaking the motion down into its horizontal and vertical components allows us to find the required solutions.

The Trajectory of Ballistic Bodies Is a Parabola

Unlike guided missiles, which follow a path which is variable and controlled by pure electronics or more sophisticated computer control systems, a ballistic body such as a football, shell, cannonball, stone or any other object projected into the air follows a parabolic trajectory after it is launched. The launching device (gun, hand, sports equipment, etc.) gives the body an acceleration, and it leaves the device with an initial velocity. The examples below ignore the effects of air drag which reduce the range and altitude attained by the body.

For lots more information on parabolas, see my tutorial: How to Understand the Equation of a Parabola, Directrix and Focus

Water from a fountain (which can be considered as a stream of particles) follows a parabolic trajectory. GuidoB, CC by SA 3.0 Unported via Wikimedia Commons

Example 1. Free-Falling Object Dropped From a Known Height

v = u + at
s= ut + ½at²
v² = u² + 2as

In this case, the falling body starts off at rest and reaches a final velocity v. The acceleration in all these problems is a = g (the acceleration due to gravity). Remember, though, that the sign of g is important, as we will see later.

Free-falling object dropped from a known height. © Eugene Brennan

Calculating final velocity

u = 0 (the body is initially at rest)

a = g (g is positive because it is in the direction of motion and accelerating the body)

s = h  (the height the object is dropped from)

The equation v = u + at can't be used because t is unknown, so use the equation v² = u²+ 2as

So:

v² = u² + 2as

= 0² + 2gh = 2gh

Taking the square root of both sides

v = √(2gh) This is the final velocity

Calculating Instantaneous Distance Fallen

s = ut + ½at²

= 0t + ½gt²

So s = ½gt²

Calculating Time Taken to Fall Distance h

s = h = ut + ½at²

= 0t + ½gt²

So h = ½gt²
Which gives

t² = 2h/g

Taking square roots of both sides

t = √(2h/g)

Calculating velocity after time t

For this we just use the equation v = u + at

So v = u + at = 0 + gt

So the velocity after an arbitrary time t is simply

v = gt

Example 2. Object Projected Vertically Upwards

v = u + at
s= ut + ½at²
v² = u² + 2as

In this scenario, the body is vertically projected upwards at 90 degrees to the ground with an initial velocity u. The final velocity v is 0 at the point where the object reaches maximum altitude and becomes stationary before falling back to Earth. The acceleration in this case is a = -g as gravity slows down the body during its upwards motion.

Object projected upwards. © Eugene Brennan

Let t₁ and t₂ be the time of flights upwards and downwards respectively

Calculating Time of Flight Upwards

v = u + at

So

0 = u + (-g)t

Giving

u = gt

So

t₁ = u/g

Calculating Distance Travelled Upwards

v² = u² + 2as

So

0² = u² + 2(-g)s

So

u² = 2gs

Giving

s = h = u²/(2g)

Calculating Time of Flight Downwards

We calculated previously that the time taken for an object to fall a distance h is:

t = √(2h/g)

But we worked out above that h =u²/(2g) is the distance travelled upwards

Substituting:

t₂ = √(2h/g) = √(2(u²/(2g))/g) = √(2u²/2g²) = u/g

Total Time of Flight

Total time of flight is t₁ + t₂ = u/g + u/g = 2u/g

Example 3. Object Projected Horizontally From a Height

v = u + at
s= ut + ½at²
v² = u² + 2as

A body is horizontally projected from a height h with an initial velocity of u relative to the ground. The key to solving this type of problem is knowing that the vertical component of motion is the same as what happens in example 1 above, when the body is dropped from a height. So as the projectile is moving forwards, it is also moving downwards, accelerated by gravity.

Object projected horizontally. © Eugene Brennan

Time of flight 

t = √(2h/g) as calculated in example one. The horizontal velocity has no bearing on how long it takes the projectile to fall.

Distance Travelled Horizontally

There is no horizontal acceleration, just a vertical acceleration g downwards  due to gravity

So distance travelled is simply velocity x time = ut = u√(2h/g)

So

s = u√(2h/g)

Example 4. Object Projected at an Angle to the Ground

v = u + at
s= ut + ½at²
v² = u² + 2as

In this example, a projectile is thrown at an angle θ to the ground with an initial velocity u. This problem is the most complex, but using basic trigonometry, we can resolve the velocity vector into vertical and horizontal components. Time of flight and vertical distance travelled to the apex of the trajectory can then be calculated using the method in example 2 (object thrown upwards). Once we have the time of flight, this allows us to calculate the horizontal distance travelled during this period.

Object projected at an angle to the ground. (The height of the muzzle from the ground has been ignored but is much less than the range and altitude.) © Eugene Brennan

A vector can be resolved into two components acting vertically and horizontally. © Eugene Brennan

Let uₕ be the horizontal component of initial velocity u.

Let uᵥ be the vertical component of initial velocity u.

So

cos θ = uₕ / u

Giving uₕ = u cos θ

Similarly

sin θ = uᵥ / u

Giving uᵥ = u sin θ

Time of Flight to Apex of Trajectory

From example 2, the time of flight is t = u/g. However since the vertical component of velocity is uᵥ

t = uᵥ /g = u sin θ/g

Altitude Attained

Again from example 2, the vertical distance travelled is s = u²/(2g) 

However since uᵥ = u sin θ is the vertical velocity:

s = uᵥ²/(2g) = (u sin θ)²/(2g))

Horizontal Distance Travelled

Since the time of flight is (usin θ)/g to the apex of the trajectory and (usin θ)/g during the period when the projectile is falling back to the ground (see downward time of flight example 2)

Total time of flight is:

(2u sin θ)/g

Now during this period, the projectile is moving horizontally at a velocity uₕ = ucos θ

So horizontal distance travelled = horizontal velocity x total time of flight

= (u cos θ) x (2u sin θ)/g

= (2u²sin θ cos θ)/g

The double angle formula can be used to simplify

I.e. sin 2A = 2sinAcosA

So (2u²sin θcos θ)/g = (u²sin 2θ)/g

Horizontal distance to apex of trajectory is half this or:

(u²sin 2θ)/2g

Recommended Books

 

Mathematics

Engineering Mathematics by KA Stroud is an excellent math textbook for both engineering students and anyone with an interest in the subject. The material has been written for part one of BSc. Engineering Degrees and Higher National Diploma courses.

A wide range of topics are covered, including some we used in this article (vectors and calculus), matrices, complex numbers, calculus applications, differential equations, series, probability theory, and statistics. The text is written in the style of a personal tutor, guiding the reader through the content, posing questions, and encouraging them to provide the answer.

I highly recommend this book. It makes learning mathematics fun!

Mechanics

Applied Mechanics by John Hanah and MJ Hillier is a standard text book for students taking Diploma and Technician courses in engineering. It covers concepts used in this article (vectors, velocity and acceleration) as well as as other topics such as motion in a circle, periodic motion, statics and frameworks, impulse and momentum, stress and strain, bending of beams and fluid dynamics. Worked examples are included in each chapter in addition to set problems with answers provided.

 

What Is the Optimum Angle to Launch a Projectile?

The optimum angle to launch a projectile is the angle which gives maximum horizontal range.

Using basic differential calculus, we can differentiate the function for horizontal range wrt θ and set it to zero allowing us to find the peak of the curve (of the graph of range versus launch angle, not the peak of the actual trajectory). Then find the angle which satisfies the equation.

So horizontal range = (u²sin 2θ)/g

Rearranging and separating out the constant gives us

u²/g (sin 2θ)

We can use the function of a function rule to differentiate sin 2θ

So if we have a function f(g), and g is a function of x, i.e. g(x)

f '(g) is the derivative of f (g) wrt g and g'(x) is the derivative of g wrt x

Then f '(x) = f '(g) g'(x)

So to find the derivative of sin 2θ, we differentiate the "outer" function giving cos 2θ and multiply by the derivative of 2θ giving 2, so

d/dθ(sin 2θ) = 2cos 2θ

Returning to the equation for range, we need to differentiate it and set it to zero to find the max range.

Using the multiplication by a constant rule

d/dθ (u²/g (sin 2θ)) = u²/g d/dθ (sin 2θ)

= u²/g (2cos 2θ)

Setting this to zero

u²/g (2cos 2θ) = 0

Divide each side by the constant 2u2/g and rearranging gives:

cos 2θ = 0

And the angle which satisfies this is 2θ = 90°

So θ = 90/2 = 45°

Orbital Velocity Formula: Satellites and Spacecraft

What happens if an objected is projected really fast from the Earth? As the object's velocity increases, it falls further and further from the point where it was launched. Eventually the distance it travels horizontally is the same distance that the Earth's curvature causes the ground to fall away vertically. The object is said to be in orbit. The velocity that this happens at is approximately 7.7 km/s or 17,000 mph in low Earth orbit.

If a body is much smaller than the object it is orbiting, the velocity is approximately:

v ≈ √(GM / r)

Where M is the mass of the larger body (in this case Earth's mass)

r is the distance from the centre of the Earth

G is the gravitational constant = 6.67430 × 10 −11 m3⋅kg−1⋅s−2

If we exceed the orbital velocity, an object will escape a planet's gravity and travel outwards from the planet. This is how the Apollo 11 crew were able to escape Earth's gravity. By timing the burn of rockets that provided propulsion and getting the velocities just right at the right moment, the astronauts were then able to insert the spacecraft into lunar orbit. Later in the mission as the LM was deployed, it used rockets to slow its velocity so that it dropped out of orbit, eventually culminating in the 1969 lunar landing.

Newton's cannonball. If the velocity is increased sufficiently, the cannonball will travel all the way around the Earth. Brian Brondel, CC by SA 3.0 via Wikipedia

A Short History Lesson

ENIAC (Electronic Numerical Integrator And Computer) was one of the first general-purpose computers designed and built during WW2 and completed in 1946. It was funded by the U.S. Army and the incentive for its design was to enable the calculation of ballistic tables for artillery shells, taking into account the effects of drag, wind and other factors influencing projectiles in flight.

ENIAC, unlike the computers of today, was a colossal machine, weighing 30 tons, consuming 150 kilowatts of power and taking up 1,800 square feet of floor space. At the time, it was proclaimed in the media as "a human brain". Before the days of transistors, integrated circuits and microprocessors, vacuum tubes (also known as "valves"), were used in electronics and performed the same function as a transistor (i.e., they could be used as a switch or amplifier).

Vacuum tubes were devices which looked like small light bulbs with internal filaments which had to be heated up with an electrical current. Each valve used a few watts of power, and since ENIAC had over 17,000 tubes, this resulted in huge power consumption. Also, tubes burnt out regularly and had to be replaced. 2 tubes were required to store 1 bit of information using a circuit element called a "flip-flop" so you can appreciate that the memory capacity of ENIAC was nowhere near what we have in computers today.

ENIAC had to be programmed by setting switches and plugging in cables and this could take weeks.

ENIAC (Electronic Numerical Integrator and Computer) was one of the first general-purpose computers. Brian Brondel, CC by SA 3.0 via Wikipedia

Vacuum tube (valve). RJB1, CC by 3.0 via Wikimedia Commons

References 

Stroud, K.A., (1970) Engineering Mathematics (3rd ed., 1987). Macmillan Education Ltd., London, England.

Hannah, J. and Hillerr, M. J., (1971) Applied Mechanics (First metric ed. 1971). Pitman Books Ltd., London, England.

Lissauer, J. J., & Pater, D. I. (2019). In Fundamental planetary science: Physics, Chemistry and Habitability (pp. 29–31). essay, Cambridge University Press.

This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualised advice from a qualified professional.

© 2014 Eugene Brennan