Answer:
Let's suppose that the hole is at y = 0m, where x is the time variable.
we know that:
The vertex is (-1s, 8m).
(i suppose x in seconds and y in meters)
At x = 1s, the ball reaches the hole, so we also have the point:
(1s, 0m).
Remember that the vertex of a quadratic equation y = a*x^2 + b*x + c is at:
x = -b/2a,
then we have:
-1 = -b/2a.
Then we have tree equations:
8m = a*(-1s)^2 + b*-1s + c
0m = a*(1s)^2 + b*1s + c
-1s = -b/2a.
First we should isolate one variable in the third equation, and then replace it in one of the other two:
1s*2a = b.
So we can replace b in the first two equations bi 1s*2a.
8m = a*1s^2 - 1s*2a*1s + c
0m = a*1s^2 + 1s*2a*1s + c
We can simplify both equations and get:
8m = a*( 1s^2 - 2s^2) + c = -a*1s^2 + c.
0m = a*(1s^2 + 2s^2) + c = a*3s^2 + c.
Easily we can isolate c in the second equation and then replace it into the first equation:
c = -a*3s^2
The first equation becomes:
8m = -a*1s^2 - a*3s^2 = -a*4s^2
a = 8m/-4s^2 = -2m/s^2.
Now with a, we can find the values of c and b.
c = -a*3s^2 = -(-2m/s^2)*3s^2 = 6m.
b = 1s*2a = 1s*(-2m/s^2) = -2m/s.
Then the equation is:
y = (-2m/s^2)*t^2 + (-2m/s)*t + 6m
(LCM is 40)
