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

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`.

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=(-b \pm √(b^2-4ac))/(2a)`.

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

`x=(-100 \pm √(100^2-4(-16)(5)))/(2(-16))`

`x=(-100 \pm √(10000+320))/(-32)`

`x=(-100 \pm √(10320))/(-32)`

`x=(-100 \pm √(16 \cdot 645))/(-32)`

`x=(-100 \pm √(16) √(645))/(-32)`

`x=(-100 \pm 4 √(645))/(-32)`

`x=((-100)/(4) \pm (4)/(4) √(645))/((-32)/(4))`

`x=(-25 \pm √(645))/(-8)`

This gives us either:

`x=(-25 + √(645))/(-8) \text{ or } (-25 - √(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.