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Introduction

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

Middle

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 from the table for example 9 then to find P we have to : f(9) =P (9²-18(9)+72=13.21

= P = 13.21/-9

= P = -1.46778(5d.p)

Conclusion

6.44/-8 = P

-0.805 = P

f (19)= P (19²-798+432)

2.89/-5= P

-0.578 = P

On the graph it occurs to be like: I predict that the curve for the last little part of the graph should be approximately around y= -0.805 (x²-42x+432). But that is just a prediction because the information given is to less to make a more precise prediction.

In conclusion, according to my investigation the equations are quadratic formulas, and using the equation y= P (x-m) (x-n) you can find the quadratic form and if you plot the values in the quadratic form then you can find value P. then after that the quadratic formula of the curve or parabola will appear. Also the value of P and M and N can be changed to change the parabola or curve.

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