Question

An amusement park ride takes ridersto the top of the tower and drops them at speeds reaching 80 ft per second. A function that models this ride is h=-16t^2-64t+60, where H is the height in feet and T is the time in seconds. About how many seconds does it take for riders to drop from 60 ft to 0 ft

Answer 1

To find the time it takes for riders to drop from 60 feet to 0 feet, we need to solve the height function \( h(t) = -16t^2 - 64t + 60 \) for the times when the height \( h(t) \) is equal to 60 feet and when it is equal to 0 feet.

1. **Finding the time when \( h(t) = 60 \)**:

We can set the height function equal to 60:

\[
-16t^2 - 64t + 60 = 60
\]

Subtracting 60 from both sides gives:

\[
-16t^2 - 64t = 0
\]

We can factor out \(-16t\):

\[
-16t(t + 4) = 0
\]

Setting each factor equal to zero:

\[
-16t = 0 \quad \Rightarrow \quad t = 0
\]

And

\[
t + 4 = 0 \quad \Rightarrow \quad t = -4
\]

Since time cannot be negative, we have \( t = 0 \).

2. **Finding the time when \( h(t) = 0 \)**:

Now, we set the function equal to 0:

\[
-16t^2 - 64t + 60 = 0
\]

To make calculations simpler, let's divide the whole equation by -4:

\[
4t^2 + 16t - 15 = 0
\]

Next, we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 16 \), and \( c = -15 \):

\[
t = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 4 \cdot (-15)}}{2 \cdot 4}
\]

Calculating the discriminant:

\[
16^2 - 4 \cdot 4 \cdot (-15) = 256 + 240 = 496
\]

Now substituting back into the formula:

\[
t = \frac{-16 \pm \sqrt{496}}{8}
\]

Calculating \( \sqrt{496} \):

\[
\sqrt{496} = \sqrt{16 \times 31} = 4\sqrt{31}
\]

Thus, we have:

\[
t = \frac{-16 \pm 4\sqrt{31}}{8} = \frac{-2 \pm \sqrt{31}}{2}
\]

This gives us two potential times:

1. \( t = \frac{-2 + \sqrt{31}}{2} \)
2. \( t = \frac{-2 - \sqrt{31}}{2} \) (which is negative and not valid for our case)

Now we will approximate \( \sqrt{31} \):

\[
\sqrt{31} \approx 5.57
\]

Substituting this back into the equation gives:

\[
t = \frac{-2 + 5.57}{2} = \frac{3.57}{2} \approx 1.785
\]

3. **Duration of the drop from 60 ft to 0 ft**:

Since the time to reach 60 ft is \( t = 0 \) and the time to reach 0 ft is approximately \( t \approx 1.785 \), the duration of the drop from 60 ft to 0 ft is:

\[
\Delta t = 1.785 - 0 = 1.785 \text{ seconds}
\]

Thus, the approximate time it takes for riders to drop from 60 feet to 0 feet is **about 1.8 seconds**.