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 ax^2 + bx^2+c = 0 that could be used to calculate when the ball is at a height of 3m. Do not solve the equation.

(-3 ÷ 4)x^2 + 6x

Step-by-step explanation:

Data mentioned in the question

Maximum height = 12m

Number of seconds = 8

Height = 3m

Depend on the above information, the quadratic equation is shown below:

As it took 8 seconds to hit the maximum altitude and it reverted to the ground floor, this graph also reflects the motion in parabola after 4 seconds, so that the a must be negative

Now it is given that

a × x ^ 2 + bx + c =0

We can considered that

x = 0

x = 8

As {0.8} are intercepts of x

When x = 0, then it is

a × 0 ^ 2 + b(0)  + c = 0 (i)

Hence 0 = 0

Now x = 8, it is

a × 8 ^ 2 + b(8)  + c = 0

Hence a(8)^2 + b(8) + c =  0 (ii)

As it can be seen that in the first equation c must be zero

Whereas the second equation is

64a + 8b = 0

i.e.

8a = -b or a = -b ÷ 8

Now according to the quadratic function, it presented

(-b ÷8)x^2 + bx + 0

So, the parabola vertex is (4, 12)

Now place this in the place of a

(-b ÷ 8)(4)^2 + b(4) = 12

And for calculating this b, all terms must be multiplied by 8

That appears

-b(16) + 32b = 96

16b = 96

So, b = 6.  

As a = -b ÷8

a = -6 ÷ 8

a = -3 ÷4

So, the equation is

= (-3 ÷ 4)x^2 + 6x

Hence, this is the equation

answer: 34rtyj,jkhtyetwrhyjeuiejnbhdfknbkguxfrbhkfdbjvkmbd,kvbsd.kxjvbhjnfbgjxdfn

step-by-step explanation: