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svetoff [14.1K]

1 year ago

7

The ball followed a path modelled by the equation h = −0.001! + 0.5 + 2.5 where h is the height of the ball in feet and is the h

orizontal distance in feet.
1) Represent the distance-height relationship for each player’s ball as an equation, in a table and on a graph. Record each representation and clearly label which player the information belongs to.

2) Supposed there were no obstacles.

a. Whose ball would travel the greatest distance before hitting the ground?

b. Whose ball would travel the shortest distance before hitting the ground?

3) Suppose the fence was 350 ft from home plate. At what height was each ball when it passed over the fence?

1 answer:

Mama L [17]1 year ago

4 0

The heights the balls hit a fence at 350 ft distance are 65 feet, 38 feet and 30 feet, respectively

<h3>Represent the distance-height relationship for each player’s ball as an equation, in a table and on a graph. </h3>

<u>Juan</u>

Juan's equation is given as:

h = -0.001d^2 + 0.5d + 2.5

h =

Set d to multiples of 50 from 0 to 400.

So, the table of values of Juan's function is:

d (ft) h(ft)

0 2.5

50 25

100 42.5

150 55

200 62.5

250 65

300 62.5

350 65

400 42.5

See attachment for the graph of Juan's function

<u>Mark</u>

A quadratic function is represented as:

h = ad^2 + bd + c

Using the values on the table of values, we have:

c = 3 -- the constant value

So, the equation becomes

h = ad^2 + bd + 3

Using the two other values on the table of values, we have:

23 = a(50)^2 + b(50) + 3

38 = a(100)^2 + b(100) + 3

Using a graphing tool, we have:

a = -0.001

b = 0.45

So, Mark's equation is h(d) = -0.001d^2 + 0.45d + 3

See attachment for Mark's graph.

<u>Barry</u>

From the graph, we have the table of values of Barry's function to be:

d (ft) h(ft)

0 2.5

50 21

100 35

150 44

200 48

250 46

300 41

350 30

400 14

450 0

Using a graphing tool, we have the quadratic function to be:

y = -0.001x^2 +0.4x +2.5

<h3><u>The shortest and the greatest distance before hitting the ground</u></h3>

From the graphs, equations and tables, the distance travelled by the balls are:

Juan = 505 feet

Mark = 457 feet

Barry = 450 feet

This means that Juan's ball would travel the greatest distance while Barry's ball would travel the shortest.

<h3>The height the balls hit a fence at 350 ft distance</h3>

To do this, we set d = 350

From the graphs, equations and tables, the height at 350 ft by the balls are:

Juan = 65 feet

Mark = 38 feet

Barry = 30 feet

The above represents the height the balls hit the fence

Read more about quadratic functions at:

brainly.com/question/12446886

#SPJ1

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We're about ready to set up an expression using those variables, but first, we should address a subtlety: the question provides us with the drainage rate in liters per second. We want the answer expressed in liters per minute, so we'll have to make that conversion beforehand. Since one second is 1/60 of a minute, a drainage rate of 400 L/s becomes 400 · 60 = 24,000 L/min.

From here, we can set up our expression. We want to find out when the tank is completely drained - when the water volume is equal to 0. If we assume that it starts full with a water volume of $V$ L, and we know that 24,000 L is drained - or subtracted - from that volume every minute, we can model our problem with the equation

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To isolate $t$ , we can take the following steps:

$V-24000t=0\\ V=24000t\\ \frac{V}{24000}=t$

So, all we need to do now to find $t$ is find $V$ . As it turns out, this is a pretty tall order. Let's begin:

Solving for $V$ :

About units: all of our measurements for the cone-shaped tank have been provided for us in meters, which means that our calculations will produce a value for the volume in cubic meters. This is a problem, since our drainage rate is given to us in liters per second. To account for this, we should find the conversion rate between cubic meters and liters so we can use it to convert at the end.

It turns out that 1 cubic meter is equal to 1000 liters, which means that we'll need to multiply our result by 1000 to switch them to the correct units.

Down to business: We begin with the formula for the area of a cone,

$V= \frac{1}{3}\pi r^2h$

which is to say, 1/3 multiplied by the area of the circular base and the height of the cone. We don't know $h$ yet, but we are given the diameter of the base: 50 m. To find the radius $r$ , we divide that diameter in half to obtain r = 50/2 = 25 m. All that's left now is to find the height.

To find that, we'll use another piece of information we've been given: a slant edge of 50 m. Together with the height and the radius of the cone, we have a right triangle, with the slant edge as the hypotenuse and the height and radius as legs. Since we've been given the slant edge (50 m) and the radius (25 m), we can use the Pythagorean Theorem to solve for the height $h$ :

$h^2+25^2=50^2\\ h^2+625=2500\\ h^2=1875\\ h=\sqrt{1875}=\sqrt{625\cdot3}=25\sqrt{3}$

With $h=25\sqrt{3}$ and $r=25$ , we're ready to solve for $V$ :

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We're almost there.

Bringing it home:

Remember that formula for $t$ we derived at the beginning? Let's revisit that. The number of minutes $t$ that it will take for this tank to drain completely is:

$t= \frac{V}{24000}$

We have our $V$ now, so let's do this:

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