For a ball on a hill, with a friction constant between the ball and the hill of $$k$$ and the acceleration due to gravity $$g$$

For an arbitrary hill $$y(x)$$ let, $$\theta = arctan({dy\over dx})$$ thus, $$g[sin(arctan({dy\over dx})) - kcos(arctan({dy\over dx}))] = a(x)$$ Equation 1 : $$a(x) = (d(v^2) \over {dx}) = 2v{dv \over {dx}} $$ Expansion : $$v = \sqrt {v_x^2 + v_y^2} = \sqrt { ({dx \over {dt}})^2 + ({dy \over {dt}})^2 } $$ Let $$dy = tan \theta dx$$ $$v = \sqrt{({dx \over {dt} })^2 + tan^2\theta({dx \over {dt}})^2}$$ $$v = {dx \over{dt}} \sqrt{1+tan^2 \theta} = {dx \over {dt}}sec \theta $$ Thus: $$v = {dx \over {dt}} \sqrt{1 + ({dy \over {dx}})^2}$$ And sub into Equation 1: $$ {d(v^2) \over {dx}} = 2{dx \over {dt}}{dv \over {dx}}\sqrt{1 + ({dy \over {dx}})^2}$$ $$v = 2 \sqrt{1 + ({dy \over {dx}})^2} {dv \over {dt}}$$ Where: $${dv \over {dt}} = a $$ Equation 2: $$a(x) = {1 \over { \sqrt {1 + ({dy \over {dx}})^2}}} {1 \over{2}}{d(v^2) \over{dx}}$$ Thus: $$ a(x) = g({1 \over { \sqrt {1 + ({dx \over {dy}})^2}}} - {k \over { \sqrt {1 + ({dy \over {dx}})^2}}}) $$ $$ a(x) = g({ { \sqrt {1 + ({dy \over {dx}})^2}} - k{ \sqrt {1 + ({dx \over {dy}})^2}} \over { { \sqrt {1 + ({dx \over {dy}})^2}} { \sqrt {1 + ({dy \over {dx}})^2}} }}) $$ $$ a(x) = g({ {dx \over {dy}}{ \sqrt {1 + ({dx \over {dy}})^2}} - k{ \sqrt {1 + ({dx \over {dy}})^2}} \over { { \sqrt {1 + ({dx \over {dy}})^2}} { \sqrt {1 + ({dy \over {dx}})^2}} }}) $$ $$ a(x) = g({{dy \over {dx}} - k \over { \sqrt {1 + ({dy \over {dx}})^2}} }) $$ And if we sub our new a(x) into 2 then: $$ { 1 \over { \sqrt {1 + ({dy \over {dx}})^2}} } { 1 \ over {2}} { d(v^2) \ over {dx}} = g({{dy \over {dx}} - k \over { \sqrt {1 + ({dy \over {dx}})^2}} }) $$ $${d(v^2) \over{dx}} = 2g({dy \over{dx}} - k) $$ Thus: $$v(x) = \pm \sqrt{2g \int{({dy \over{dx}} -k )dx}}$$ and we know that the length of any line(arc) of any function is given by, $$ \int { \sqrt{ 1 + ({dy\over dx})^2 } }dx = d(x)$$ so lets review what we now have,

$$d(x) = \int { \sqrt{ 1 + ({dy\over dx})^2 } }dx$$ $$v(x) = \pm \sqrt{2g \int{({dy \over{dx}} -k )dx}}$$ $$ a(x) = g({{dy \over {dx}} - k \over { \sqrt {1 + ({dy \over {dx}})^2}} }) $$ and from $$d = vt$$ $${d(x)\over v(x)} = t(x)$$ so $$t(x) = {{\int { \sqrt{ 1 + ({dy\over dx})^2 } }dx} \over {\pm \sqrt{2g \int{({dy \over{dx}} -k )dx}}} }$$ so for two hills

Velocity as a function of x

Time as a function of x

Velocity as a function of x

Time as a function of x

Our project will examine how identical balls rolling down differently shaped slopes of the same height behave with regard to final speed and travel time. The balls will likely be steel bearings and they will roll down hills made of wood or or plexiglass. The fastest ball and the final velocities of the balls will be measured and reported. Each ball will have two sensors at the bottom of their track, and a release mechanism at the top to ensure simultaneous release. The fastest ball will be the first ball to trip the first sensor on its track, and the speed of each ball will be measured using the two sensors of a known separation. The total time taken for each ball to roll down the hill will be measured from the time the ball is released, until it hits the first sensor at which point the travel time will be recorded.

## Doug's Responsiblities

- Programming
- Calculations
- Powering the Game

## Kurtis' Responsiblities

- Construction of Game
- Measurement Accuracy
- User Interface