Question

Kira is a lovable dog who is full of energy. Her owner thought it would be fun to train her by throwing a frisbee for her to catch. When the frisbee is thrown, it follows a parabolic path that is modeled by the function h(t) = – 0.145t2 + 0.019t + 5.5. How many seconds will it take for the frisbee to hit the ground?

–6.2 seconds
–6.1 seconds
5.5 seconds
6.2 seconds

Answer 1

To find out when the frisbee hits the ground, we need to determine when the height \( h(t) \) equals zero. This means we need to solve the equation:

\[
h(t) = -0.145t^2 + 0.019t + 5.5 = 0
\]

This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where:

- \( a = -0.145 \)
- \( b = 0.019 \)
- \( c = 5.5 \)

Using the quadratic formula:

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

Let's plug in the values for \( a \), \( b \), and \( c \):

1. Calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = (0.019)^2 = 0.000361
\]
\[
4ac = 4 \cdot (-0.145) \cdot 5.5 = -3.19
\]

Now, the discriminant is:

\[
0.000361 - (-3.19) = 0.000361 + 3.19 = 3.190361
\]

2. Now we compute \( t \):

\[
t = \frac{-0.019 \pm \sqrt{3.190361}}{2 \cdot -0.145}
\]

Calculating \( \sqrt{3.190361} \):

\[
\sqrt{3.190361} \approx 1.786
\]

Now, substituting back:

\[
t = \frac{-0.019 \pm 1.786}{-0.29}
\]

Now we will calculate both possible values for \( t \):

1. For \( t_1 = \frac{-0.019 + 1.786}{-0.29} \):

\[
t_1 = \frac{1.767}{-0.29} \approx -6.09 \quad (\text{not a valid time since it's negative})
\]

2. For \( t_2 = \frac{-0.019 - 1.786}{-0.29} \):

\[
t_2 = \frac{-1.805}{-0.29} \approx 6.21
\]

Rounding to one decimal place, we get \( t \approx 6.2 \) seconds.

Thus, the frisbee will hit the ground at approximately **6.2 seconds**. Therefore, the correct answer is:

**6.2 seconds**.