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

Respuesta :

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 [tex]y=-16x^2+100x+5[/tex].

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, [tex]x[/tex], such that the height, [tex]y[/tex], is 0.

We are going to solve the following equation:

[tex]y=-16x^2+100x+5[/tex] with [tex]y=0[/tex]

[tex]0=-16x^2+100x+5[/tex]

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

[tex]a=-16[/tex]

[tex]b=100[/tex]

[tex]c=5[/tex]

The quadratic formula is:

[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex].

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

[tex]x=\frac{-100 \pm \sqrt{100^2-4(-16)(5)}}{2(-16)}[/tex]

[tex]x=\frac{-100 \pm \sqrt{10000+320}}{-32}[/tex]

[tex]x=\frac{-100 \pm \sqrt{10320}}{-32}[/tex]

[tex]x=\frac{-100 \pm \sqrt{16 \cdot 645}}{-32}[/tex]

[tex]x=\frac{-100 \pm \sqrt{16} \sqrt{645}}{-32}[/tex]

[tex]x=\frac{-100 \pm 4 \sqrt{645}}{-32}[/tex]

[tex]x=\frac{\frac{-100}{4} \pm \frac{4}{4} \sqrt{645}}{\frac{-32}{4}}[/tex]

[tex]x=\frac{-25 \pm \sqrt{645}}{-8}[/tex]

This gives us either:

[tex]x=\frac{-25 + \sqrt{645}}{-8} \text{ or } \frac{-25 - \sqrt{645}}{-8}[/tex]

Let's punch both of these into the calculator:

[tex]x \approx -0.04961 \text{ or } 6.299601[/tex]

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