######
53

An object was shot up into the air at an initial vertical speed of \(320\) feet per second. Its height as time passes can be modeled by the quadratic function \(f\text{,}\) where \(f(t)={-16t^{2}+320t}\text{.}\) Here \(t\) represents the number of seconds since the object’s release, and \(f(t)\) represents the object’s height in feet.

After , this object reached its maximum height of .

This object flew for before it landed on the ground.

This object was in the air \({12\ {\rm s}}\) after its release.

This object was \({1584\ {\rm ft}}\) high at two times: once after its release, and again later after its release.

######
54

An object was shot up into the air at an initial vertical speed of \(384\) feet per second. Its height as time passes can be modeled by the quadratic function \(f\text{,}\) where \(f(t)={-16t^{2}+384t}\text{.}\) Here \(t\) represents the number of seconds since the object’s release, and \(f(t)\) represents the object’s height in feet.

After , this object reached its maximum height of .

This object flew for before it landed on the ground.

This object was in the air \({7\ {\rm s}}\) after its release.

This object was \({1520\ {\rm ft}}\) high at two times: once after its release, and again later after its release.

######
55

From a clifftop over the ocean \({200\ {\rm m}}\) above sea level, an object was shot into the air with an initial vertical speed of \({156.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) On its way down it fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function \(f\text{,}\) where \(f(t)={-4.9t^{2}+156.8t+200}\text{.}\) Here \(t\) represents the number of seconds since the object’s release, and \(f(t)\) represents the object’s height (above sea level) in meters.

After , this object reached its maximum height of .

This object flew for before it landed in the ocean.

This object was above sea level \({26\ {\rm s}}\) after its release.

This object was \({748.8\ {\rm m}}\) above sea level twice: once after its release, and again later after its release.

######
56

From a clifftop over the ocean \({160\ {\rm m}}\) above sea level, an object was shot into the air with an initial vertical speed of \({176.4\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}\) On its way down it fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function \(f\text{,}\) where \(f(t)={-4.9t^{2}+176.4t+160}\text{.}\) Here \(t\) represents the number of seconds since the object’s release, and \(f(t)\) represents the object’s height (above sea level) in meters.

After , this object reached its maximum height of .

This object flew for before it landed in the ocean.

This object was above sea level \({8\ {\rm s}}\) after its release.

This object was \({1742.7\ {\rm m}}\) above sea level twice: once after its release, and again later after its release.

######
57

A remote control aircraft will perform a stunt by flying toward the ground and then up. Its height can be modeled by the function \(h(t)={1.2t^{2}-16.8t+55.8}\text{.}\) The plane

hit the ground during this stunt.######
58

A remote control aircraft will perform a stunt by flying toward the ground and then up. Its height can be modeled by the function \(h(t)={0.1t^{2}-1.6t+10.4}\text{.}\) The plane

hit the ground during this stunt.######
59

A submarine is traveling in the sea. Its depth can be modeled by \(d(t)={-0.9t^{2}+16.2t-72.9}\text{,}\) where \(t\) stands for time in seconds. The submarine

hit the sea surface along this route.######
60

A submarine is traveling in the sea. Its depth can be modeled by \(d(t)={-1.6t^{2}+28.8t-133.6}\text{,}\) where \(t\) stands for time in seconds. The submarine

hit the sea surface along this route.######
61

An object is launched upward at the height of \(400\) meters. It’s height can be modeled by

\begin{equation*}
h=-4.9t^2+90t+400\text{,}
\end{equation*}

where \(h\) stands for the object’s height in meters, and \(t\) stands for time passed in seconds since its launch. The object’s height will be \(420\) meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be \(420\) meters. Round your answers to two decimal places if needed.

The object’s height would be \(420\) meters the first time at seconds, and then the second time at seconds.

######
62

An object is launched upward at the height of \(210\) meters. It’s height can be modeled by

\begin{equation*}
h=-4.9t^2+70t+210\text{,}
\end{equation*}

where \(h\) stands for the object’s height in meters, and \(t\) stands for time passed in seconds since its launch. The object’s height will be \(250\) meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be \(250\) meters. Round your answers to two decimal places if needed.

The object’s height would be \(250\) meters the first time at seconds, and then the second time at seconds.

######
63

Currently, an artist can sell \(220\) paintings every year at the price of \({\$60.00}\) per painting. Each time he raises the price per painting by \({\$5.00}\text{,}\) he sells \(5\) fewer paintings every year.

Assume he will raise the price per painting \(x\) times, then he will sell \(220-5x\) paintings every year at the price of \(60+5x\) dollars. His yearly income can be modeled by the equation:

\begin{equation*}
i=(60+5x)(220-5x)
\end{equation*}

where \(i\) stands for his yearly income in dollars. If the artist wants to earn \({\$18{,}375.00}\) per year from selling paintings, what new price should he set?

To earn \({\$18{,}375.00}\) per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.

######
64

Currently, an artist can sell \(250\) paintings every year at the price of \({\$90.00}\) per painting. Each time he raises the price per painting by \({\$15.00}\text{,}\) he sells \(5\) fewer paintings every year.

Assume he will raise the price per painting \(x\) times, then he will sell \(250-5x\) paintings every year at the price of \(90+15x\) dollars. His yearly income can be modeled by the equation:

\begin{equation*}
i=(90+15x)(250-5x)
\end{equation*}

where \(i\) stands for his yearly income in dollars. If the artist wants to earn \({\$37{,}125.00}\) per year from selling paintings, what new price should he set?

To earn \({\$37{,}125.00}\) per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.