Ball motion with Sage

Jose Guzman(*) and Ignacio Delgado

version 1.0.1

December 09, 2009

(*) This file is under construction. Please send any suggestions or comments to [email protected]

Index

1. Introduction:

We will consider the equation to describe the vertical motion of a ball, and analyze and solve with Sage. Code is extensively commented through the notebook and should be self-explanatory. Comments and suggestions are greatly appreciated.

2. Ball motion equation:

From Newton's second law of motion we can write a mathematical formulation of a ball motion. With this equation we can calculate the ball's vertical position (y) according to the time as follows:

$y(t) = V_ot -\frac{1}{2}gt^2$

where $y$ is the vertical distance, $V_o$ is the initial velocity of the ball, $g$ is the acceleration of gravity, and $t$ is time.

# define equation
var('t,v0,g') # symbolic variables
y(t)=v0*t-(g*t**2)/2 # ball motion equation

#result = y.subs(v0=5,g=9.81,t=1)
#print result.pyobject() # we need pyobject() to get the value

3. Solving the equation with Sage:

We can look for for solutions to the equation a y=0, to know the time it takes for the ball to move upwards and returns to y again.

$V_ot -\frac{1}{2}gt^2 =0 ; t(V_o -\frac{1}{2}gt) = 0$

We have two possible solutions, corresponding to

$t=0$ or $t=2V_o/g$

We can easily solve this with Sage.

solve(y==0,t) # solve equation y=0 with respect to t

[t == 2*v0/g, t == 0]

We can calculate the maximal vertical distance reached by simply taking the derivative of the function and solve it against time.

${\displaystyle\frac{dy}{dt}=V_o -gt}$

Once Again, Sage allows us to perform this simple calculation.

diff(y,t) # calculate dy/dt

t |--> -g*t + v0

If we equal the equation to zero.

${\displaystyle\frac{dy}{dt}=V_o -gt=0 :}$

we can calculate the time at which the maximum vertical distance is reached

$t=V_o/g$

In Sage, we simply solve the differential equation with respect to t.

solve(diff(y,t)==0,t)

[t == v0/g]

4. Graphic representation:

We can plot it now all together.

import numpy as np

class Hyperbola():
    
    def __init__(self, velocity, color):
        """ 
        creates a hyperbolic curve of the
        ball motion at a given initial velocity.
         
        Arguments:
        velocity    -- initial velocity
        color       -- color to plot the curve 
        """
        # public variables
        #f(v) = v*t - (g*t**2)/2
        v,t=var('v,t') # symbolic variables
        g=9.81

        self._v = velocity
        self._color = color
        self.f = velocity*t - (g*t**2)/2 # this only depends on t now  
        self._tend = 2*velocity/g
        self.__g=9.81
        self._tmax = self._v/self.__g

    def maxcoor(self):
        """
        returns x,y coordinates of the point at the maximum
        """ 
        ymax = self.f(self._tmax)
        return (self._tmax, ymax)

    def plot(self):
        """ 
        creates a plot object with the curve, label, maximun point
        and the coordinates of the max point
        """
        curve = plot(self.f,0,self._tend,rgbcolor=self._color)
        circle = point(self.maxcoor(), rgbcolor='white', pointsize=30,faceted=True)

        location = (self._tmax,self.f(self._tmax)*1.2)
        label = text("V=%s m/s"%self._v, location,rgbcolor=self._color)

        return curve + circle + label


# plot functions
# create some objects
myfunc = Hyperbola(velocity=6,color='orange')
#print myfunc.plot()
#print myfunc.maxcoor()
figure = myfunc.plot()
v=[2,3,4,5] # list of velocities
colors = ['red','green','black','blue'] # list of colors
for i,j in zip(v,colors):
    myfun = Hyperbola(velocity=i, color=j)
    figure += myfun.plot()


figure.fontsize(14)
figure.show(xmax=1.50)