Question

A model rocket is launched with an initial upward velocity of 100 f/s. The rocket's height h (in feet) after t seconds is given by the following.
H=100t-16t^2
Find all values of t for which the rocket's height is 34 feet.
Round your answers) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

To find the values of \( t \) for which the rocket's height \( H \) is 34 feet, we start with the given equation for height:

\[
H = 100t - 16t^2
\]

We need to set \( H \) equal to 34 feet:

\[
34 = 100t - 16t^2
\]

Rearranging this equation gives us:

\[
16t^2 - 100t + 34 = 0
\]

Next, we can apply the quadratic formula, which is given by:

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

In our case, \( a = 16 \), \( b = -100 \), and \( c = 34 \). First, we can calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-100)^2 - 4 \cdot 16 \cdot 34
\]
\[
b^2 - 4ac = 10000 - 2176
\]
\[
b^2 - 4ac = 7824
\]

We can use this in the quadratic formula to find \( t \):

\[
t = \frac{-(-100) \pm \sqrt{7824}}{2 \cdot 16}
\]
\[
t = \frac{100 \pm \sqrt{7824}}{32}
\]

Calculating \( \sqrt{7824} \):

\[
\sqrt{7824} \approx 88.49
\]

Substituting back, we get:

\[
t = \frac{100 \pm 88.49}{32}
\]

This yields two values for \( t \):

1.
\[
t_1 = \frac{100 + 88.49}{32} = \frac{188.49}{32} \approx 5.89
\]

2.
\[
t_2 = \frac{100 - 88.49}{32} = \frac{11.51}{32} \approx 0.36
\]

Therefore, the two values of \( t \) for which the height of the rocket is 34 feet are:

\[
t \approx 0.36 \text{ seconds or } t \approx 5.89 \text{ seconds}
\]

So, the final rounded answers are:

**t ≈ 0.36 or t ≈ 5.89**.