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Dovator [93]
3 years ago
11

Scott sets up a volleyball net in his backyard. One of the poles, which forms a right angle with the ground, is 12 feet high. To

secure the pole, he attached a rope from the top of the pole to a stake 10 feet from the bottom of the pole. Find the length of the rope, rounded to the nearest tenth.

Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

Length of the rope = 15.6 ft.

Step-by-step explanation:

Given:

Height of the pole from the ground = 12 ft

Adjacent distance from the pole = 10 ft

We have to find the length of the rope attached by Scott to secure the pole.

According to the question :

In forma  right angled triangle as shown in the figure.

We have to find the measure of the hypotenuse.

Length of rope = Hypotenuse measure

Let the length of the rope = "h" ft

Using Pythagoras formula.

⇒ (hypotenuse)=\sqrt{opposite^2+ adjacent^2}

⇒ Plugging the values.

⇒ h=\sqrt{12^2 + 10^2}

⇒ h=\sqrt{144 + 100}

⇒ h=\sqrt{244}

⇒ h=15.62 ft

The length of the rope, rounded to the nearest tenth is 15.6 ft.

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a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

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