On a distant planet, a ball is thrown upwards from ground level , reaching a maximum height of 12m and hitting the ground again in eight seconds. Determine a quadratic equation in the form a * x ^ 2 + bx + c =0 that could be used to calculate when the ball is a height of 3m. Do not solve the equation

(-3/4)x^2 + 6x = 3

Step-by-step explanation:

It takes 8 seconds for the ball to reach its maximum height and then return to ground level; thus, the highest point (vertex) the ball will reach is reached after 4 seconds.  The graph showing this motion is a parabola that opens down.  Thus, the coefficient 'a' must be negative.

Returning to a * x ^ 2 + bx + c, we write out two instances of this formula, one for x = 0 and the other for x = 8.  This is because {0, 8} are the x-intercepts.

Then, for x = 0, a * x ^ 2 + bx + c   becomes  a * 0 ^ 2 + b*0 = 0,

and for x = 8,  a * x ^ 2 + bx + c  becomes  a(8)^2 + b(8) + c = 0

The first equation tells us that 'c' must be 0.  The second equation tells us that  64a + 8b = 0, which is equivalent to 8a = -b, or a = -b/8.

We then have the quadratic function  (-b/8)x^2 + bx + 0.  All we have to do now is to find the value of b.  The vertex of this parabola is (4, 12), and so the quadratic becomes

(-b/8)(4)^2 + b(4) = 12.  To solve this for b, multiply all terms by 8:

-b(16) + 32b = 96, or

16b = 96, or b = 6.  Since a = -b/8 (see above), a = -6/8 or a = -3/4.

Then the equation of motion is

height of ball = (-3/4)x^2 + 6x

We want to know when the ball is at a height of 3 m.  Thus, set this equation of motion = to 3 and solve for x (the times at which height = 3 m):

(-3/4)x^2 + 6x = 3            This is the desired quadratic equation.

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