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Solving Projectile Motion Problems - Applying Newton's Equations of Motion to Ballistics

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Physics, Mechanics, Kinematics and Ballistics

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 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 Newton's equations of motion.

See my hub Force, Weight, Newtons, Velocity, Mass and Friction - Basic Principles of Mechanics which introduces the basic concepts of mechanics

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

Newton's Equations of Motion - Formulas for Distance and Velocity

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 body has an initial velocity u, and after time t, it reaches a velocity v. It also travels a distance s.

So we have 5 parameters associated with the body in motion: u, v, a, s and t

Newton's equations of motion allow us to work out any of these parameters once we know three other parameters. So the three formulae are:

v = u + at

s= ut + ½at2

v2 = u2 + 2as


Projectile Motion Problems - Calculating Time of Flight, Distance Traveled and Altitude

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

There are 4 scenarios normally presented in these types of problems, and it is necessary to calculate parameters mentioned above:

  1. Object dropped from a known altitude
  2. Object thrown upward
  3. Object thrown horizontally from a height above the ground
  4. Object launched from the ground at an angle

These problems are solved by considering the initial conditions and this allows to work out a formula for velocity, distance travelled, time of flight etc. In example 3 and 4, 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 shell, cannon ball, particle or stone thrown into the air follows a parabolic trajectory. The examples below ignore the effects of air drag which reduce the range and altitude attained by the body.

Water from a fountain (which can be considered as a stream of particles) follows a parabolic trajectory
Water from a fountain (which can be considered as a stream of particles) follows a parabolic trajectory | Source

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.

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)

So:

v2 = u2 + 2as

= 02 + 2gh = 2gh

Taking the square root of both sides

v = √(2gh) This is the final velocity


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)



Free Falling Object dropped from a known height
Free Falling Object dropped from a known height | Source

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.


Calculating time of flight upwards


v = u + at

So

0 = u + (-g)t

Giving

u = gt

So

t = u/g

Calculating time of flight downwards

This is also u/g. You can calculate it knowing the altitude attained as worked out below and knowing that the initial velocity is zero. Hint: use example 1 above!


Total time of flight

total time of flight is u/g upwards + u/g downwards = 2u/g

Calculating distance traveled upwards

v2 = u2 + 2as

So

02 = u2 + 2(-g)s

So

u2 = 2gs

Giving

s = h = u2/(2g)


Object projected upwards
Object projected upwards | Source

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

Time of flight


t = √(2h/g) as calculated in example 1


Distance traveled horizontally

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

So distance traveled = velocity x time = ut = u√(2h/g)

So

s = u√(2h/g)


Object projected horizontally
Object projected horizontally | Source

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 traveled 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 traveled during this period.

See diagram below

Let uh be the horizontal component of initial velocity

Let uv be the vertical component of initial velocity

So

Cos Θ = uh / u

Giving uh = uCos Θ

Similarly

Sin Θ = uv / u

Giving uv = uSin 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 = uSin Θ/g


Altitude attained

Again from example 2, the vertical distance traveled is s = u2/(2g). However since uv = uSin Θ is the vertical velocity:

s = uv2/(2g) = (uSin Θ)2/(2g))


Horizontal distance traveled

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:

2uSin Θ/g

Now during this period, the projectile is moving horizontally at a velocity uh = uCos Θ

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

= uCos Θ x 2uSin Θ/g

= 2u2Sin ΘCos Θ/g



A Vector Can be Resolved Into Two Components

Vertical and horizontal components of velocity
Vertical and horizontal components of velocity | Source
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)
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) | Source

Optimum Angle to Launch a Projectile

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

Using basic 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 actual trajectory). Then find the angle Θ which satisfies the equation.

So horizontal range = 2u2Sin ΘCos Θ/g

Derivative of Sin Θ is Cos Θ

Derivative of Cos Θ is -Sin Θ

Using the product rule and isolating the constant 2u2/g :

d/dΘ(2u2Sin ΘCos Θ/g)

= 2u2 /g [(Sin Θ)(-Sin Θ) + (Cos Θ)( Cos Θ)]

Setting this to zero

2u2 /g [(Sin Θ)(-Sin Θ) + (Cos Θ)( Cos Θ)] = 0

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

(Sin Θ)2 = (Cos Θ)2

So Sin Θ = Cos Θ

And the angle which satisfies this is Θ = 45ο

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 kilowatt of power and taking up 1800 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 micropressors, 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 of 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
ENIAC (Electronic Numerical Integrator And Computer) was one of the first general purpose computers | Source
Vacuum tube (valve)
Vacuum tube (valve) | Source

Engineering Mathematics by K.A. Stroud

A groundbreaking and comprehensive reference with over 500,000 copies sold since it first debuted in 1970, the new seventh edition of Engineering Mathematics has been thoroughly revised and expanded. An interactive Personal Tutor CD-ROM is included with every book. Providing a broad mathematical survey, this innovative volume covers a full range of topics from the very basic to the advanced.

I highly recommend this book as it's like having a personal tutor who guides you through the solution of problems as it progresses through the text and explains everything clearly.

1 comment

ojhadv 2 years ago

its quite a refreshing article to read back class 7 physics again,

I had always used to throw a cricket ball to calculate its speed depending on distance after i learnt these equations, made physics interesting

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