Question

y=-16x^2+100x+5) since it is being launched from a platform 5 feet high and has an initial velocity of 100 ft/sec. Find the time when the rocket crashes to the ground

Answer 1

Answer:

The one that makes more sense for our conclusion is that the rocket crashes approximately 6.299601 seconds after it has been launched given the path of the rocket is $y=-16x^2+100x+5$.

Step-by-step explanation:

The rocket has crashed on the ground when the height between the ground and the rocket is 0.

We want to find the time, $x$, such that the height, $y$, is 0.

We are going to solve the following equation:

$y=-16x^2+100x+5$ with $y=0$

$0=-16x^2+100x+5$

Upon comparing this equation to $ax^2+bx+c=0$, I see that I have the following values for $a,b,$ and $c$:

$a=-16$

$b=100$

$c=5$

The quadratic formula is:

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

Let's plug in the values we got above now.

$x=\frac{-100 \pm \sqrt{100^2-4(-16)(5)}}{2(-16)}$

$x=\frac{-100 \pm \sqrt{10000+320}}{-32}$

$x=\frac{-100 \pm \sqrt{10320}}{-32}$

$x=\frac{-100 \pm \sqrt{16 \cdot 645}}{-32}$

$x=\frac{-100 \pm \sqrt{16} \sqrt{645}}{-32}$

$x=\frac{-100 \pm 4 \sqrt{645}}{-32}$

$x=\frac{\frac{-100}{4} \pm \frac{4}{4} \sqrt{645}}{\frac{-32}{4}}$

$x=\frac{-25 \pm \sqrt{645}}{-8}$

This gives us either:

$x=\frac{-25 + \sqrt{645}}{-8} \text{ or } \frac{-25 - \sqrt{645}}{-8}$

Let's punch both of these into the calculator:

$x \approx -0.04961 \text{ or } 6.299601$

The one that makes more sense for our conclusion is that the rocket crashes approximately 6.299601 seconds after it has been launched.