Re: paper: Parabolic Trajectory Calculations
 

Re: paper: Parabolic Trajectory Calculations
 

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.

Re: paper: Parabolic Trajectory Calculations
 

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)

Re: paper: Parabolic Trajectory Calculations
 

You are misunderstanding the meaning of terminal velocity.

Re: paper: Parabolic Trajectory Calculations
 

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.

Re: paper: Parabolic Trajectory Calculations
 

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.

Re: paper: Parabolic Trajectory Calculations
 

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.

Re: paper: Parabolic Trajectory Calculations
 

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