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

Step-by-step explanation:

Data provided in the question

Maximum height = 12m

Number of seconds = 8

Height = 3m

Based on the above information, the quadratic equation is as follows

Since it took 8 seconds for reaching the maximum height and then it returned to the ground level so here the highest point is done after 4 seconds also this graph represents the motion in parabola so the a should be negative

Now it is mentioned that

a × x ^ 2 + bx + c =0

We can assume that

x = 0

x = 8

As these {0.8} are intercepts of x

When x = 0, then it would be

a × 0 ^ 2 + b(0)  + c = 0 (i)

Therefore 0 = 0

Now x = 8, it would be

a × 8 ^ 2 + b(8)  + c = 0

Therefore a(8)^2 + b(8) + c =  0 (ii)

As we can see  that in the first equation c should be zero

While the second equation would be

64a + 8b = 0

i.e.

8a = -b or a = -b ÷ 8

Now as per the quadratic function, it appears

(-b ÷8)x^2 + bx + 0

Now the parabola vertex is (4, 12)

Now put this in the place of a

(-b ÷ 8)(4)^2 + b(4) = 12

Now for solving this b, all terms should be multiplied by 8

That comes

-b(16) + 32b = 96

16b = 96

So, b = 6.  

As a = -b ÷8

a = -6 ÷ 8

a = -3 ÷4

Now the equation is

= (-3 ÷ 4)x^2 + 6x

Hence, this is the equation

The

Might be 300, im not sure. Hopefully it's right.