Constant Acceleration Model

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Constant acceleration model (without coordinates!!):

\[\Large{\frac{{d{\bf{v}}}}{{dt}}\,\,\, = \,\,\,{\bf{g}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\bf{v}}\,\, = \,\,\,{{\bf{v}}_0}\,\, + \,\,{\bf{g}}t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{d{\bf{x}}}}{{dt}}\,\,\, = \,\,\,{\bf{v}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\bf{x}}\,\,\, = \,\,\,{{\bf{x}}_0}\,\, + \,\,{{\bf{v}}_0}t\,\, + \,\,\frac{1}{2}{\bf{g}}{t^2}}\]

                               (hodograph)                                      (trajectory)

v ½ g t 2 ½ g t ½ g t /t r r 0 v t 0 v g t 2 /t r 0 v 0 v v v

Vector algebraic model
r = range = x - x0):

\[\Large{\bbox[8px,border:2px solid red]{\frac{1}{2}\left( {{\bf{v}} + {{\bf{v}}_0}} \right)\,\,\, = \,\,\,\frac{{\bf{r}}}{t}\,\,\, \equiv \,\,\,{\bf{\bar v}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{v}} - {{\bf{v}}_0}\,\,\, = \,\,\,{\bf{g}}t}}\]

          Reduces all projectile problems to "
solving a parallelogram!!"

Problem: Determine 

          (a)  the range r of a target sighted in a direction  r/r  that has been hit by a projectile,
                launched with velocity  

          (b)  the launching angle for maximum range, 

          (c)  the time of flight.


          General case: Elevated target.

                    Complicated solution with rectangular coordinates published in AJP. 

                    Much simpler GA solution in NFCM (see references in Prolog).

                    (Click here for solutions for this problem.)

          Simplest case: r is horizontal. 

                    Hint: Consider (v0)(v0)

                    (Partial solution follows below on this page.


Solving a parallogram with GA:          − v0 gt                    v0 2r/t

g t 2 /t r 0 v 0 v v v

          We eliminate t by multiplication:    (v + v0)(v - v0= 2rg

          Expanding:    v2 − v0v0v − vv0 v2 − v0+ 2v0v = 2(rrg

          Separately equating scalar and bivector parts, we get:

                              v2 − v02 2rg                    v0v rg

          The first equation may be recognized as conservation of energy (loss in KE = gain in PE).

          The second equation represents a geometry theorem:

                    If a parallelogram has area A,
                    a new parallelogram, made with the diagonals of the first, has area 2

For r horizontal, we have 

g t 2 /t r 0 v 0 v v v θ θ

\[\Large{\begin{array}{l}{\bf{r}} \cdot {\bf{g}}\,\,\, = \,\,\,0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{v^2}\,\,\, = \,\,\,{v_0}^2\,\,\,\,\,\, \Rightarrow \\\left| {{{\bf{v}}_0} \wedge {\bf{v}}} \right|\,\,\, = \,\,\,{v_0}^2\,\,\sin \left( {2\theta } \right)\,\,\, = \,\,\,\left| {{\bf{r}} \wedge {\bf{g}}} \right|\,\,\, = \,\,\,rg\end{array}}\]

Hence, the
horizontal range formula:

\[\Large{\bbox[8px,border:2px solid red]{\begin{array}{l}r\,\,\, = \,\,\,\frac{{{v_0}^2}}{g}\sin \left( {2\theta } \right)\end{array}}}\]

The general GA solution for an elevated target and time of flight is derived in

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Next section: Circular Motion Model.

© David Hestenes 2005, 2014