Question

A rocket is launched from the top of aa 5050 foot cliff with an initial velocity of 150150 feet per second. The​ height, h, of the rocket after t seconds is given by the equation h equals negative 16 t squared plus 150 t plus 50h=−16t2+150t+50. How long after the rocket is launched will it be 1010 feet from the​ ground?
Question content area bottom
Part 1
talmost equals≈enter your response here
▼

Answer 1

To find how long after the rocket is launched it will be 10 feet from the ground, we need to set the height equation to 10 feet and solve for \( t \).

The height equation given is:

\[
h(t) = -16t^2 + 150t + 50
\]

We set \( h(t) \) equal to 10:

\[
10 = -16t^2 + 150t + 50
\]

Rearranging this equation gives:

\[
0 = -16t^2 + 150t + 50 - 10
\]

\[
0 = -16t^2 + 150t + 40
\]

To make the computation simpler, we can multiply the entire equation by -1:

\[
0 = 16t^2 - 150t - 40
\]

Now we can apply the quadratic formula to find \( t \):

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

Here, \( a = 16 \), \( b = -150 \), and \( c = -40 \).

Calculating the discriminant \( b^2 - 4ac \):

\[
b^2 = (-150)^2 = 22500
\]
\[
4ac = 4 \cdot 16 \cdot -40 = -2560
\]
\[
b^2 - 4ac = 22500 - (-2560) = 22500 + 2560 = 25060
\]

Now substituting the values into the quadratic formula:

\[
t = \frac{-(-150) \pm \sqrt{25060}}{2 \cdot 16}
\]

\[
t = \frac{150 \pm \sqrt{25060}}{32}
\]

Next, we calculate \( \sqrt{25060} \):

\[
\sqrt{25060} \approx 158.25
\]

Substituting this back into the equation for \( t \):

\[
t = \frac{150 \pm 158.25}{32}
\]

Calculating the two possible times:

1. \( t = \frac{150 + 158.25}{32} = \frac{308.25}{32} \approx 9.64 \)
2. \( t = \frac{150 - 158.25}{32} = \frac{-8.25}{32} \approx -0.26 \) (not valid since time cannot be negative)

Thus, the valid solution for the time when the rocket is 10 feet from the ground is approximately:

\[
t \approx 9.64 \text{ seconds}
\]

Therefore, the final answer is:

\[
\mathbf{t \approx 9.64}
\] seconds.