Question

Please Help I understand how to start but not how to finish

Answer 1

Answer: 10.97 c

Explanation:

y=-16x²+170x+61

When the rocket completes its flight, its height of flight will be y=0 ft.

`Hence,\ 0=-16x^2+170x+61`

Multiply both parts of the equation by -1:

`0=16x^2-170x-61\\`

`Thus,\ 16x^2-170x-61=0`

`\displaystyle\\a=16\ \ \ \ \ b=-170\ \ \ \ c=-61\\\\D=b^2-4ac\\\\D=(-170)^2-4(16)(-61)\\\\D=28900+3409\\\\D=32804\\\\√(D) =√(32804) \\\\√(D) =181,1187\\\\x=(-bб√(D) )/(2(a)) \\\\x=(-(-170)б181,1187)/(2(16)) \\\\x=(170б181,1187)/(32)\\\\ x=-0.34746\\otin(x\geq 0)\\\\x=10.972\\\\x\approx10.97\ c`

Answer 2

Given:

`y=-16x^2_{}+170x+61`

Where x represents the time in seconds and y represents the height.

To find the time that the rocket will hit the ground:

Put y=0,

`-16x^2_{}+170x+61=0`

Using the quadratic formula,

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-170\pm\sqrt[]{(170)^2-4(-16)(61)}}{2(-16)} \\ =\frac{-170\pm\sqrt[]{28900+3904}}{-32} \\ =\frac{-170\pm\sqrt[]{32804}}{-32} \\ =(-170\pm181.118)/(-32) \\ x=(-170+181.118)/(-32),x=(-170-181.118)/(-32) \\ x=-0.3474,10.9724 \\ x\approx-0.347,10.972 \end{gathered}`

Hence, the time required to hit the ground in 10.972 seconds.