IGCSE Modeling Project bouncing ball

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IGCSE Modeling Project bouncing ball

The aim of this investigation is to find out the equation that is suitable for the parabola on the graph. According to my experiment I find out that using a quadratic formula can use be suitable for parabola. The following equation proves my analysis:

                                        f(x)= ax²+bx+c ( this is a quadratic formula, this is the final                                                                        equation that we will get for the parabola)

For the first parabola the equation is y= -1.725x²+10.02x+1.99

This equation if you type it into a autograph program it will draw a parabola that lies on the points of the first bounce:


to find the values for the equation I use this formula:

                                                                                                        = P (x-m) (x-n)

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                                                                                                        = P (x²-nx-mx+mn)

                                                                                                        = P (x²+x(-n-m)+mn)

This formula will turn to a quadratic formula after solving it, and then it would be the equation for the parabola.

For example for the second parabola if you plot in the lowest and highest X coordinate in the bounce it should come out like this: = P (x-6) (x-12)

Then if you solve the equation it should turn out like: y= P (x²-18x+72)

Value P basically is the value that influence the curve so it’s a very important value in this equation. In order to find out P we have to take point ...

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