Question

a man throws a ball off the top of a building and records the height of the ball at different times as shown in the table:

Height of the ball

time (s) Height (feet)
0 46
1 63
2 48
3 1

a. Find a quadratic model for the data
b. Use the model to estimate the height of the ball at 2.5 seconds
c. What is the ball's maximum height?

Answer 1

Okay, so first of all the original quadratic equation is in this form:
\[ ax ^{2} + bx + c = y(x)\]
let "x" represent your time and "y" as you height.

we then pick points on the table, for example let's pick when time is "0" the height is "46" feet in which we use as x(time) = 0 and y(height) = 46 and then use them in our equation:
\[a(0)^{2} + b(0) + c = 46\]
so we see here that "c" = 46

Now lets choose two more point values: (1, 63) and (3,1) and then plug into our equation:
\[a(1)^{2} + b(1) + 46 = 63\]
lets call this Equation #1

then second point value
\[a(3)^2 +b(3) + 46 =1\] which simplifies to\[9a + 3b + 46 =1\] and call this Equation #2
Our job here is to find what "a" and "b" are since we have already found out what "c" is.
So solving for Equation #1 for b leads us to this:\[b = 17-a\]
you then plug that into Equation #2 to eliminate "b" so we can solve for "a" alone:
\[9a + 3(17-a) +46 = 1\]

From here you solve for "a". After finding "a" plug your value back into either equation and then solve for "b". After you have all three values plug them back into your original equation:
\[ax ^{2} + bx + c = y(x) \]
You will have your quadratic equation for part A.
To solve for part B, simply plug in the given time where the "x" is, since after all, that is your time spot.

Part C should be easily seen from the table with the highest number as your max height.