Question

A model rocket is launched with an initial upward velocity of 188 ft/s. The rocket's height h (in feet) after t seconds is given by the following.
H=188t-16t^2
Find all values of t for which the rocket's height is 92 feet.

Answer 1

To find the values of \( t \) for which the rocket’s height \( h(t) \) is 92 feet, we can set the height equation equal to 92:

\[
H = 188t - 16t^2
\]

We need to solve the equation:

\[
188t - 16t^2 = 92
\]

Rearranging it gives:

\[
-16t^2 + 188t - 92 = 0
\]

To make it easier to work with, we can multiply the entire equation by -1:

\[
16t^2 - 188t + 92 = 0
\]

Now we will use the quadratic formula to find the values of \( t \). The quadratic formula is:

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

For our equation, \( a = 16 \), \( b = -188 \), and \( c = 92 \).

Now, let's calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = (-188)^2 = 35344
\]
\[
4ac = 4 \cdot 16 \cdot 92 = 7392
\]
\[
b^2 - 4ac = 35344 - 7392 = 27952
\]

Now we can find the square root of the discriminant:

\[
\sqrt{27952} \approx 167.2
\]

Now we can substitute back into the quadratic formula:

\[
t = \frac{-(-188) \pm 167.2}{2 \cdot 16}
\]

Calculating the two possible values for \( t \):

1. For the plus sign:
\[
t_1 = \frac{188 + 167.2}{32} = \frac{355.2}{32} \approx 11.1
\]

2. For the minus sign:
\[
t_2 = \frac{188 - 167.2}{32} = \frac{20.8}{32} \approx 0.65
\]

Thus, the two values of \( t \) for which the rocket's height is 92 feet are approximately:

\[
t \approx 0.65 \text{ seconds and } t \approx 11.1 \text{ seconds.}
\]