Trajectory Position At A Given Time

AI Thread Summary
To model projectile motion and determine the X and Y coordinates at any given time, the discussion emphasizes using idealized conditions such as a flat earth, constant gravity, and no air resistance. The equations derived from Newton's laws provide a straightforward solution for the horizontal and vertical positions over time. The horizontal position is calculated using x(t) = v_x(0)t, while the vertical position is given by y(t) = Y_0 + v_y(0)t - (g t^2)/2, where v_y(0) is the initial vertical velocity. For more complex scenarios, numerical approximation methods may be necessary. Understanding these foundational equations allows for accurate modeling of projectile motion in graphics programs.
sklar
Messages
1
Reaction score
0
I am writing a graphics program to model projectile motion of a ball being thrown. What I am looking for is an equation that will give a specific X and Y coordinate, in the parabolic path, for any specific time it is given, and a arbitratry velocity and angle.

For example, if I give it some arbitrary velocity and angle, I would like to be able to figure out how high in the air the ball is and how far it has traveled horizontally at 5 seconds, 13.4 seconds or any other random time.

Any help would be appreciated.

Sklar
 
Physics news on Phys.org
What sort of simplifying assumptions are you making? A flat earth, constant acceleration of gravity w.r.t. altitude, no air drag? Since you mention "parabolic path," I imagine you are going with all of these idealizations.

If you idealize (simplify) things enough, you can make use of closed-form solutions that were worked out with calculus centuries ago. If you are going to make it more realistic, you will probably have to resort to numerical approximation methods.
 
A strictly time depentent solution is pretty easy. The only force acting is gravity. Application of Newtons laws gives.

in the x direction

v_x (0) =v_i(0) cos(\theta)
so
x(t) = v_x(0)t + C

If you define the origion as the point where the ball is thrown C=0

Gravity acts in the y direction so

\frac {d^2y} {dt^2} = -g

\frac {dy} {dt} = -gt + v_y(0)
but
v_y(0) = v_i(0) sin (\theta)

y(t) = Y_0 + v_y(0)t - \frac {g t^2} 2
 
Last edited:
And thereby Integral has derived your parabolic curve, parameterized by time t.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top