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Marina CMI [18]
3 years ago
5

Super Dave Osborne stands at the bottom of a canyon elevation 256 feet below sea level, he throws a piano up at a velocity of 3

20 feet per second the piano lands on the edge of the cliff at zero elevation . The flight of the piano is modeled by the equation f( x ) = −16t2 2 + 320t − 256. The piano was in the air for exactly how many seconds
Mathematics
1 answer:
fgiga [73]3 years ago
6 0

Answer:

The piano was in the air for 18.34 seconds.

Step-by-step explanation:

The height of the piano after t seconds is given by the following equation:

f(t) = -16t^{2} + 320t - 256

Initially, the piano is below the ground level. At the first root of the equation, it will go into the air, and then some time after that it will fall down again, which is at the second root of the equation.

Roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

f(t) = -16t^{2} + 320t - 256

So

a = -16, b = 320, c = -256

\bigtriangleup = (320)^{2} - 4*(-16)*(-256) = 86016

t_{1} = \frac{-320 + \sqrt{86016}}{2*(-16)} = 0.8348

t_{2} = \frac{-320 - \sqrt{86016}}{2*(-16)} = 19.17

Subtracting:

19.17 - 0.8348 = 18.34

The piano was in the air for 18.34 seconds.

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Sally Sue had spent all day preparing for the prom. All the glitz and the glamour of the evening fell apart as she stepped out o
Nuetrik [128]

We have the function

p(t)=550(1-e^{-0.039t})

Therefore we want to determine when we have

p(t_0)=550

It means that the term

e^{-0.039t}

Must go to zero, then let's forget the rest of the function for a sec and focus only on this term

e^{-0.039t}\rightarrow0

But for which value of t? When we have a decreasing exponential, it's interesting to input values that are multiples of the exponential coefficient, if we have 0.039 in the exponential, let's define that

\alpha=\frac{1}{0.039}

The inverse of the number, but why do that? look what happens when we do t = α

e^{-0.039t}\Rightarrow e^{-0.039\alpha}\Rightarrow e^{-1}=\frac{1}{e}

And when t = 2α

e^{-0.039t}\Rightarrow e^{-0.039\cdot2\alpha}\Rightarrow e^{-2}=\frac{1}{e^2}

We can write it in terms of e only.

And we can find for which value of α we have a small value that satisfies

e^{-0.039t}\approx0

Only using powers of e

Let's write some inverse powers of e:

\begin{gathered} \frac{1}{e}=0.368 \\  \\ \frac{1}{e^2}=0.135 \\  \\ \frac{1}{e^3}=0.05 \\  \\ \frac{1}{e^4}=0.02 \\  \\ \frac{1}{e^5}=0.006 \end{gathered}

See that at t = 5α we have a small value already, then if we input p(5α) we can get

\begin{gathered} p(5\alpha)=550(1-e^{-0.039\cdot5\alpha}) \\  \\ p(5\alpha)=550(1-0.006) \\  \\ p(5\alpha)=550(1-0.006) \\  \\ p(5\alpha)=550\cdot0.994 \\  \\ p(5\alpha)\approx547 \end{gathered}

That's already very close to 550, if we want a better approximation we can use t = 8α, which will result in 549.81, which is basically 550.

Therefore, we can use t = 5α and say that 3 people are not important for our case, and say that it's basically 550, or use t = 8α and get a very close value.

In both cases, the decimal answers would be

\begin{gathered} 5\alpha=\frac{5}{0.039}=128.2\text{ minutes (good approx)} \\  \\ 8\alpha=\frac{8}{0.039}=205.13\text{ minutes (even better approx)} \end{gathered}

7 0
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