Plot 2D Trajectory: Solve Initial Velocity & Coordinates

Click For Summary
To plot the trajectory of an object starting at the origin (0,0), the correct equation for the y-coordinate is y = 100*t - 4.9*t^2. By setting t = 0, the initial point is correctly calculated as y0 = 0. The user was mistakenly using an initial velocity equation that began at y = 100, leading to an incorrect starting point of (0, 95.1). Adjusting the time variable allows for the accurate plotting of the trajectory as a parabola. This ensures the trajectory starts at the desired coordinates.
rd42
Messages
6
Reaction score
0
I'm trying to plot out the trajectory of an object in a program that is looping through the function and spitting out the x and y coordinates, but I'm having trouble. I want to see a parabola starting at 0,0 going up and ending again at x,0.

Basic question:

(units aren't critical here f/s m/s anything as long as they're all the same)
initial velocity in x = 50
initial velocity in y = 100

x coordinate:
x = v * t
x = 50 * 1
x1 = 50

y coordinate:
y = v + .5gt^2
y = 100 + -4.9
y1 = 95.1

I'm missing something here. How do I get the points along the path of the trajectory starting at 0,0 instead of starting at 0, 95.1 as the first coordinate?

Thanks,
rd42
 
Last edited:
Physics news on Phys.org
rd42 said:
y coordinate:
y = v + .5gt^2
y = 100 + -4.9
y1 = 95.1

I'm missing something here. How do I get the points along the path of the trajectory starting at 0,0 instead of starting at 0, 95.1 as the first coordinate?

Thanks,
rd42

You mean y = 100*t -4.9*t2

You get point y1 when you set t =1 in the above equation. You get the earlier point y0 when you set t = 0 in the above equation.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
22K
  • · Replies 1 ·
Replies
1
Views
11K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
5
Views
5K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 0 ·
Replies
0
Views
612