Imagine separating a parabola into its different x and y components and linking them by the component of time.(adsbygoogle = window.adsbygoogle || []).push({});

Scenario:

Given:

Gravity

No Drag

The velocity of the launched object is not sufficient to escape the gravitational pull.

Ignoring:

The change of gravity as the distance between the centre of the two masses increases (projectile and planet constitute as the two masses)

An object is launched on a flat plain, at an angle of Θ, with a velocity of Ω.

There are two components to this parabola; the constant x velocity, the changing y velocity (due to gravity)

The x velocity is

X=(Ω(90-Θ))⁄90

This equation splits the initial velocity into x and y values. Imagine if you launched an object strait up, there is no x velocity. If an object is launched at a gradient of tan45 (tan is equivalent to rise over run) then the initial velocity will be evenly divided between x and y.

Note x is constant

The x distance:

Xdist=time(( Ω(90-Θ))⁄(90 ))

The y velocity:

The y velocity is constantly changing due to the nature of gravity.

The initial y velocity is

Y=(Ω(Θ))⁄90

And now gravity is taken into account

Velocity of falling object:

V= Velocity

A = Acceleration constant

T= Time

V=AT

Gravity’s pull is a negative addition to the y velocity.

Therefore the overall y velocity can be summarised as:

Yvel=(Ω(Θ))⁄(90 )-AT

We don’t want velocities; we need distances to find the equation of the parabola.

A

To find the equation for this graph one would find the distance equation,

V=at

If this is integrated (D=distance)

D=1/2 at^2

Therefore the overall equation for the y height is:

YHeight= -1/2 at^2+ t(Ω(Θ)⁄(90 ))

Xdist=time(( Ω(90-Θ))⁄(90 ))

Given that the x distance is related to the y distance by time we can combine the equations to get the parabola that is formed. Or alternately 3 time points can be substituted into the above x and y equations along with all the other information required, then one can solve simultaneously for the equation. The constant (c) in the general form of the quadratic equation y=ax^2+bx+c can be found, it is equal to the height of launch.

The next step is to factorise in drag, into the model of the flight path of the projectile, unfortunately at this point parabolas cannot model the flight path of the object.

The quadratic drag equation

Circia 1900’s, Lord Rayleigh

Fd=-1/2 pv^2 ACdv

Where

Fd= force of drag, p = the density of the fluid, v=velocity, A= reference area,

Now

Consider.

A force instantaneous affair, given this if a relationship to time can be realised from this equation, and it can then be integrated, the sum total of the entire negative vector up until a point can calculate giving the velocity at the corresponding time point.

One of the properties of air resistance is that due to its non linear nature the drag must be calculated on the total velocity, and then separated into its x and y components.

The velocity of the object is:

V=( Ω(90-Θ))⁄(90 ))*p +(Ω(Θ))⁄(90 )-AT*p

V= velocity

P=launch velocity

this is as far as i have gone.

How can drag be added to the different x and y components with a link to time?

note: all this has been done with year 11 maths and i could be completely wrong, this is not homework, this is just out of interest. how can the path of a rocket which has an non linear (change in gravity and air resistance) accelerating velocity. non linear changing angle of flight respective to (center of gravity, velocity, angle of launch, and much more ) be modeled.

my teacher thinks that the answer will be a differential equation.

thank you to whoever replies!

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# Factoring drag into artillery flight, differential equation

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