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djyliett [7]
3 years ago
10

A toy rocket is shot vertically into the air from a launching pad 9 feet above the ground with an initial velocity of 168 feet p

er second. The height h, in
feet, of the rocket above the ground as t seconds after launch is given by the function h(t)=-16t^2+168t+9. How long will it take the rocket to reach its maximum height? What is the maximum height?
Mathematics
2 answers:
Lana71 [14]3 years ago
8 0

The maximum or minimum of a function is the point
where its first derivative = 0.

                             h(t) = -16t² + 168t + 9

First derivative  h'(t) =  -32t + 168
 
                   h'(t) = 0  =   -32t + 168 = 0

Add 32t to each side:     32t = 168

Divide each side by 32:     <em>t = 5.25 seconds</em>

                              
                                                 h(t) = -16t² + 168t + 9

Height at 5.25 seconds =        -16(5.25²) + 168(5.25) + 9 = 

                                                     -16(27.5625)   +   882    +    9 =

                                                                - 441      +    882   +  9  =<em>  450-ft </em>

hammer [34]3 years ago
5 0
Y(initial) = 9
V(initial) = 168
V(final) = 0
g(accel-grav) = 32 (in feet per second
squared)
Use the following equation:
V(final) = V(initial) + a(t)
Since this object is moving straight up and
down, a = -32
Enter the knowns into the equation
0 = 168 - 32t
32t = 168
t = 168/9.8
t = 5.25

Now that you know the time it takes to
reach it's maximum, use the general
kinematics equation to solve for final
distance:

y(final) = y(initial) + V(initial)(t) + (1/2)a
(t²)

y(final) = 9 + 168(5.25) + (1/2)(-32)
(5.25²)

= 9 + 882 - 441

= 450 feet
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Solution:

Step 1:

We will calculate the volume of ice cream in the single scoop

The volume of the ice cream will be

\begin{gathered} V=\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3 \\ r=\frac{2in}{2}=1in(cone) \\ h=4.5in \\ r=\frac{3in}{2}=1.5in(radius\text{ of the hemisphere\rparen} \end{gathered}

By substituting the values, we will have

\begin{gathered} V=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^{3} \\ V=\frac{1}{3}\times\frac{22}{7}\times1^2\times4.5+\frac{2}{3}\times\frac{22}{7}\times1.5^3 \\ V=\frac{33}{7}+\frac{99}{14} \\ V=\frac{165}{14} \\ V=11.79in^3 \end{gathered}

Step 2:

We will use the formula below to calculate the volume of the two scoops of ic cream

\begin{gathered} V=\frac{1}{3}\pi r^2h+\frac{4}{3}\pi r^3 \\ V=\frac{1}{3}\times\frac{22}{7}\times1^2\times4.5in+\frac{4}{3}\times\frac{22}{7}\times1.5^3 \\ V=\frac{33}{7}+\frac{99}{7} \\ V=\frac{132}{7} \\ V=18.86in^3 \end{gathered}

Step 3:

We will use the formula below to calculate the volume of the three scoops of ic cream

\begin{gathered} V=\frac{1}{3}\pi r^2h+\frac{6}{3}\pi r^3 \\ V=\frac{1}{3}\times\frac{22}{7}\times1^2\times4.5+2\times\frac{22}{7}\times1.5^3 \\ V=\frac{33}{7}+\frac{297}{14} \\ V=\frac{363}{14} \\ V=25.93in^3 \end{gathered}

For the first ice cream with one scoop

\begin{gathered} 1in^3=\frac{3.50}{11.79} \\ 1in^3=\text{ \$}0.30 \end{gathered}

For the second ice cream with two scoops

\begin{gathered} 1in^3=\frac{4.50}{18.86} \\ 1in^3=\text{ \$}0.24 \end{gathered}

For the third ice cream with three scoops

\begin{gathered} 1in^3=\frac{5.50}{25.93} \\ 1in^3=\text{ \$}0.21 \end{gathered}

Hence,

The final answer is

The triple sold at $5.50 has the best value because it has the lowest price of $0.21 per cubic inch of the ice cream

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