paper: Parabolic Trajectory Calculations

[GeeTwo]

Originally Posted by peronis

does anyone now the terminal velocity of the fuel?

Search is your friend.

[Ether]

Originally Posted by Jacob Plicque

The terminal velocity can be calculated from the peak height and the gravity constant.

OK, so given g=-9.8 meters/sec² and peak height = 3 meters, please show us how you would calculate the terminal velocity... without using any other information such as launch speed or launch angle.

[Jacob Plicque]

From a known height of 3 meters the calculation follows a freefall model (Vy=0 @t=0) which uses Vy=gt and Y=0.5gt^2.
Thus t=sqrt(3m/9.8m/sec^2/0.5sec)=0.78 seconds.
Vy terminal=0.78sec*9.8m/sec^2=7.66m/sec
The Newtonian trajectory equations do use the initial velocity Voy as follows:
Vy=Voy-gt
Y=Voyt-0.5gt^2 @ Vy=0 the arc is at its peak
X=Voht
Voy=Vo*sin(launch angle from horizon)
Vox=Vo*cos(launch angle)

[Ether]

Originally Posted by Jacob Plicque

From a known height of 3 meters the calculation follows a freefall model (Vy=0 @t=0) which uses Vy=gt and Y=0.5gt^2.
Thus t=sqrt(3m/9.8m/sec^2/0.5sec)=0.78 seconds.

You are misunderstanding the meaning of terminal velocity.

[Jacob Plicque]

The Newtonian model equations ignore drag which calculates a velocity only on the basis of gravity. My answer is only valid for a drag free Newtonian calculation. Terminal velocity with drag provide an upper velocity limit for a falling object. For a falling baseball it would be about 33m/sec (~108ft/sec) while a hail stone is ~14m/sec (45ft/sec). A 3 meter drop reaches a peak velocity of 7.66m/sec (~26ft/sec) which is less than the maximum terminal velocity. of either of the above examples.

[Ether]

Originally Posted by Jacob Plicque

The Newtonian model equations ignore drag which calculates a velocity only on the basis of gravity. My answer is only valid for a drag free Newtonian calculation. Terminal velocity with drag provide an upper velocity limit for a falling object. For a falling baseball it would be about 33m/sec (~108ft/sec) while a hail stone is ~14m/sec (45ft/sec). A 3 meter drop reaches a peak velocity of 7.66m/sec (~26ft/sec) which is less than the maximum terminal velocity. of either of the above examples.

The definition of terminal velocity in the context of this thread is given in the first paragraph of this web page:

https://en.wikipedia.org/wiki/Terminal_velocity

Please read it.

You are using a different definition.

[Ether]

Originally Posted by Jacob Plicque

The Newtonian model equations ignore drag which calculates a velocity only on the basis of gravity.

Models with air drag are "Newtonian" too. The acceleration is still equal to the net force divided by the mass (Newton's 2nd law).

Perhaps what you meant is constant acceleration model equations ignore air drag.

[Ether]

Originally Posted by Ether

Thread created automatically to discuss a document in CD-Media.

Parabolic Trajectory Calculations by Ether

I've recently received a couple of PMs asking about the formulas in the spreadsheet. I'm going to post summaries of the answers here so interested students may benefit:

Some cells are hidden to make the user interface cleaner. To make those cells visible: Unhide columns GHIJK, move the graph out of the way, and highlight the whole spreadsheet.

The air drag acceleration vector D always points 180 degrees opposite to the Velocity vector V.

The magnitude of D is modeled as:

(V²/V_t²)*g

... where V_t is the magnitude of the terminal velocity.


The magnitude of the X component of D is

(V²/V_t²)*g*cos(θ)

= (V²/V_t²)*g*(Vx/V)

= (V*V_x/V_t²)*g


The magnitude of the Y component of D is

(V*V_y/V_t²)*g