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3 years ago

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The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

object is 150 feet.
After how many seconds does the object reach its maximum height?

2 answers:

Monica [59]3 years ago

7 0

namely what is "t" when h(t) is 150, so

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+96t+6\implies \stackrel{h(t)}{150}=-16t^2+96t+6\implies 16t^2-96t+144=0 \\\\\\ t^2-6t+9=0\implies (t-3)(t-3)=0\implies \boxed{t=3}$

Margarita [4]3 years ago

7 0

Answer:

3 seconds

Step-by-step explanation:

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Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

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Now, from equation (i)

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This is the required area function in the terms of variable $x$ .

For the domain of area function, from equation (ii)

$x=15-\frac{2y}{3}$

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For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

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Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

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