A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.

Question

y = - 16x ^ 2 + 118x + 69

GPT-4o mini · Answer

To find the time when the rocket will hit the ground, we need to determine when the height $y$ is equal to 0. This means we need to solve the equation:

$- 16 x^{2} + 118 x + 69 = 0$

We can use the quadratic formula to solve for $x$ , given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Here, $a = - 16$ , $b = 118$ , and $c = 69$ . First, we need to calculate the discriminant:

$b^{2} - 4 a c = 118^{2} - 4 \cdot (- 16) \cdot 69$

Calculating $b^{2}$ :

$118^{2} = 13924$

Calculating $4 a c$ :

$4 \cdot (- 16) \cdot 69 = - 4256$

Now we can compute the discriminant:

$b^{2} - 4 a c = 13924 - (- 4256) = 13924 + 4256 = 18180$

Next, we apply the quadratic formula:

$x = \frac{- 118 \pm \sqrt{18180}}{2 \cdot - 16}$

Calculating $\sqrt{18180}$ :

$\sqrt{18180} \approx 134.86$

Now substituting this back into the formula:

$x = \frac{- 118 \pm 134.86}{- 32}$

Calculating the two possible values for $x$ :

First value:

$x_{1} = \frac{- 118 + 134.86}{- 32} = \frac{16.86}{- 32} \approx - 0.5275 (not a valid solution, as time cannot be negative)$

Second value:

$x_{2} = \frac{- 118 - 134.86}{- 32} = \frac{- 252.86}{- 32} \approx 7.91$

Thus, the time that the rocket will hit the ground, to the nearest hundredth of a second, is:

$7.91 seconds$