Question

Jimmy threw a baseball in the air from the roof of his house. The path followed by the baseball can be modeled by the function f ( t ) = − 8 t 2 + 48 t + 40 , f ( t ) = - 8 t 2 + 48 t + 40 , where t represents the time in seconds after the ball was thrown and f ( t ) f ( t ) represents its height, in feet, from the ground. Part A. How high is the roof from the ground? How many seconds did it take for the ball to hit the ground after it was thrown off the roof? Part B. Jimmy wanted to throw the ball at a maximum height of 120 feet. Did Jimmy's baseball reach this height after it was thrown? Explain your answer.

Answer 1

Answer:

Step-by-step explanation:

The first part of A is easy. Look at the quadratic function, and the constant, the very last number with no t stuck to it represents the height from which the object in question was originally launched. Our constant is 40, so the height of the roof from which the baseball was thrown is 40 feet. Part 2 of A is not quite as simple because it requires factoring using the quadratic formula.Before we do that, let's make our numbers a bit more manageable, shall we? Let's factor out a -8 to get

$f(t) = -(t^2-6t-5)$ and a = 1, b = -6, c = -5.

Filling in the quadratic formula now looks like this:

$t=\frac{6+-\sqrt{6^2-4(1)(-5)} }{2(1)}$ and

$t=\frac{6+-\sqrt{36+40} }{2}$ and

$t=\frac{6+-\sqrt{56} }{2}$ so the 2 solutions are

$t=\frac{6+\sqrt{56} }{2}=6.74sec$ and

$t=\frac{6-\sqrt{56} }{2}=-.742sec$ and since we know time can NEVER be negative, the time it takes for the baseball to hit the ground from a height of 40 feet is 6.74 seconds. Onto part B.

In order to determine exactly how high the baseball did go, we have to find the vertex of the function. We do this by completing the square and getting the function into vertex, or work, form. Begin by setting the quadratic equal to 0, moving over the constant, and then factoring out the leading coefficient. The rule for completing the square are kinda picky in that you have to have a 1 as the leading coefficient, and righ now ours is a -8. So following the rules I stated above:

$-8(t^2-6t)=-40$ Next is the take half the linear term, square it, and then add it to both sides. Our linear term is a -6. Half of -6 is -3, and -3 squared is 9, so we add 9 into the parenthesis first:

$-8(t^2-6t+9)=-40+??$

Because this is an equation, we can't add 9 to one side without adding the equivalent to the other side. But, we cannot forget about that -8 sitting out front there, refusing to be ignored. We didn't just add in a 9, we actually added in a -8 times 9 which is -72. That's what goes on the right side in place of the ??.

$-8(t^2-6t+9)=-40-72$

The reason we complete the square is found on the left side of the equals sign. We have, in the process of completing the square, formed a perfect square binomial that will serve as the h in our vertex (h, k) where h is the number of seconds it takes for the baseball to reach its max height of k, whatever k is. That's what we have to find out. Putting the left side into its simplified perfect square binomial and adding the numbers on the right gives us:

$-8(t-3)^2=-112$

For the last step, add over the -112 and set it back equal to f(t):

$-8(t-3)^2+112=f(t)$ From that we determine that the vertex is (3, 112). The max height of this baseball was 112 feet...so no, it did not make it up to the height of 120 feet that Jimmy wanted for the baseball.