projectile motions 1

8/3/2019 Projectile Motions 1

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In physics, assuming a flat Earth with a uniform gravityfield, a projectilelaunched with specific initial

conditions will have a predictable range. As in Trajectory of a projectile, we will use:

The following applies for ranges which are small compared to the size of the Earth. For longer rangessee sub-orbital spaceflight.

g: the gravitational accelerationusually taken to be 9.80 m/s2 (32 f/s2) near the Earth's surface

: the angle at which the projectile is launched

v: the velocity at which the projectile is launched

y0: the initial height of the projectile

d: the total horizontal distance travelled by the projectile

When neglecting air resistance, the range of a projectile will be

If (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the

projectile will then simplify to

Ideal projectile motion

Ideal projectile motion assumes that there is no air resistance. This assumption simplifies the math

greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are

small. Ideal projectile motion is also a good introduction to the topic before adding the complications of air

resistance.

Derivations

A) Flat Ground

First we examine the case where (y0) is zero. The horizontal position (x(t)) of the projectile is

In the vertical direction

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We are interested in the time when the projectile returns to the same height it originated at,

thus

By factoring:

or

The first solution corresponds to when the projectile is first launched. The second solution is the useful

one for determining the range of the projectile. Plugging this value for (t) into the horizontal equation

yields

Applying the trigonometric identity

sin(x+ y) = sin(x)cos(y) + sin(y)cos(x)

If x and y are same,

sin(2x) = 2sin(x)cos(x)

allows us to simplify the solution to

Note that when () is 45, the solution becomes

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B) Uneven Ground

Now we will allow (y0) to be nonzero. Our equations of motion are now

and

Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how

we defined our starting height to begin with)

Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic

manipulation

The square root must be a positive number, and since the velocity and the cosine of the launch angle can

also be assumed to be positive, the solution with the greater time will occur when the positive of the plus

or minus sign is used. Thus, the solution is

Solving for the range once again

Maximum Range

For cases where the projectile lands at the same height from which it is launched, the maximum range isobtained by using a launch angle of 45 degrees. A projectile that is launched with an elevation of 0

degrees will strike the ground immediately (range = 0), though it may then bounce or roll. A projectile that

is fired with an elevation of 90 degrees (i.e. straight up) will travel straight up, then straight down, and

strike the ground at the point from which it is launched, again yielding a range of 0.

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The elevation angle which will provide the maximum range when launching the projectile from a non-zero

initial height can be computed by finding the derivative of the range with respect to the elevation angle

and setting the derivative to zero to find the extremum:

where and R = horizontal range.

Setting the derivative to zero provides the equation:

Substituting u = (cos)2 and 1 u = (sin)2 produces:

Which reduces to the surprisingly simple expression:

Replacing our substitutions yields the angle that produces the maximum range for uneven ground,

ignoring air resistance:

Note that for zero initial height, the elevation angle that produces maximum range is 45 degrees, as

expected. For positive initial heights, the elevation angle is below 45 degrees, and for negative initial

heights (bounded below by y0 > 0.5v2 / g), the elevation angle is greater than 45 degrees.

Example: For the values g= 9.81m / s2,y0 = 40m , and v = 50m / s, an elevation angle = 41.1

produces a maximum range ofRmax = 292.1m.

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Actual projectile motion

In addition to air resistance, which slows a projectile and reduces its range, many other factors also have

to be accounted for when actual projectile motion is considered.

Projectile characteristics

Generally speaking, a projectile with greatervolume faces greater air resistance, reducing the range of

the projectile. This can be modified by the projectile shape: a tall and wide, but short projectile will face

greater air resistance than a low and narrow, but long, projectile of the same volume. The surface of the

projectile also must be considered: a smooth projectile will face less air resistance than a rough-surfaced

one, and irregularities on the surface of a projectile may change its trajectory if they create more drag on

one side of the projectile than on the other. Massalso becomes important, as a more massive projectile

will have more kinetic energy, and will thus be less affected by air resistance. The distribution of mass

within the projectile can also be important, as an unevenly weighted projectile may spin undesirably,

causing irregularities in its trajectory due to the magnus effect.

If a projectile is given rotation along its axesof travel, irregularities in the projectile's shape and weight

distribution tend to be canceled out. See rifling for a greater explanation.

Firearm barrels

For projectiles that are launched by firearms and artillery, the nature of the gun's barrel is also important.

Longer barrels allow more of the propellant's energy to be given to the projectile, yielding greater

range. Rifling, while it may not increase the average (arithmetic mean) range of many shots from the

same gun, will increase theaccuracy and precision of the gun.

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