Question

Suppose the height of a punt (in feet) after t seconds can be modeled by the function

Answer 1

Answer:

a) 51 feet

b) 3.5 seconds

Step-by-step explanation:

The y-coordinate of the vertex of the given parabola is what we are looking for.

We first need to find the t-coordinate of the vertex.

The t-coordinate can be found using -b/(2a).

We need to compare

-16t^2+56t+2

to

at^2+bt+c

to identify a,b, and c.

a=-16

b=56

c=2

We are ready to compute -b/(2a).

-b/(2a)=-56/(2*-16)=-56/-32=7/4.

The vertex occurs at t=7/4.

To find y, we use y=2+56t-16t^2

y=2+56(7/4)-16(7/4)^2

y=51

So the maximum height is 51 feet.

Part b)

Hitting ground means the height between the ball and the ground is 0.

So we need to replace h(t) with 0.

0=2+56t-16t^2

I'm going to use quadratic formula.

a=-16

b=56

c=2

The quadratic formula is:

$t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$t=\frac{-56 \pm \sqrt{56^2-4(-16)(2)}}{2(-16)}$

Computing the thing inside the square root and the thing in the denominator using my handy dandy calculator:

$t=\frac{-56 \pm \sqrt{3264}}{-32}{/tex]I'm going to do the square root of 3264 now:[tex]t=\frac{-56 \pm 57.131427428}{-32}$.

I'm going to compute both

$\frac{-56 + 57.131427428}{-32}$ and $-56-57.131427428}{-32}$ using my handy dandy calculator:

$-0.035357$ while the other one is $3.535357$

The negative value doesn't make sense for our problem so the answer is approximately 3.5 seconds.