Question

Every year in Delaware there is a contest where people create cannons and catapults designed to launch pumpkins as far in the air as possible. the equation y=12+105x-16x^2 can be used to represent the height,y, of a launched pumpkin, where x is the time in seconds that the pumpkin has been in the air. What is the maximum height that the pumpkin reaches? How many seconds have passed when the pumpkin hits the ground? (Hint if the pumpkin hits the ground, its height is 0 feet.)
A: The pumpkin's maximum height is 184.27 feet and it hits the ground after 3.28 seconds.
B: The pumpkin's maximum height is 3.28 feet and it hits the ground after 6.67 seconds.
C: The pumpkin's maximum height is 184.27 feet and it hits the ground after 6.67 seconds
D: The pumpkin's maximum height is 3.28 feet and it hits the ground after 184.27 seconds

Answer 1

Answer:  C. The pumpkin's maximum height is 184.27 feet and it hits the ground after 6.67 seconds

Step-by-step explanation:

Since, the given function that shows the height,

$f(x) = 12 + 105 x - 16x^2$  ------- (1)

Where x is the time in seconds.

Since, the given function is a parabola,

Therefore the maximum height of the parabola is found at the vertex.

And, the vertex of the parabola = $\frac{-2b}{a}$

Where a is the coefficient of $x^2$ and b is the coefficient of x.

Thus, the vertex of the given parabola = $\frac{-2\times 105}{-16}=3.281$

Therefore, the maximum height of the pumpkin = $f(3.281)= 12 + 105\times 3.281 - 16\times (3.281)^2 = 184.266$ feet ≈ 184.27 feet

Now, the pumpkin hits the ground when f(x)= 0,

⇒ $12 + 105 x - 16x^2=0$

⇒ x = 6.675 seconds ≈ 6.67 seconds

Thus, Option C is correct.

Answer 2

My numbers aren't exactly the same due to rounding but you can still see the answer.