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Cloud [144]
3 years ago
13

Jake attempts a 36-yard field goal in a football game. For his attempt to be a success, the football needs to pass through the u

prights and over the crossbar that is 10 feet above the ground. Jake kicks the ball from the ground with an initial velocity of 68 feet per second, at an angle of 34° with the horizontal. What is true of Jake's attempt?
The kick is good! The ball passes about 10 feet above the crossbar.

The kick is good! The ball clears the crossbar by approximately 14 feet.

The kick is good! The ball clears the crossbar by about 4 feet.

The kick is not successful. The ball is 2 feet too low.
Mathematics
1 answer:
Arturiano [62]3 years ago
7 0

Question:

1) The kick is good! The ball passes about 10 feet above the crossbar.

2) The kick is good! The ball clears the crossbar by approximately 14 feet.

3) The kick is good! The ball clears the crossbar by about 4 feet.

4) The kick is not successful. The ball is 2 feet too low.

Answer:

The correct option is;

3) The kick is good! The ball clears the crossbar by about 4 feet.

Step-by-step explanation:

Here we have

Horizontal distance from goal, x = 36 yards = 108 ft

Vertical distance of goal, y = 10 ft

Angle of elevation of ball = 34°

Initial velocity of ball, v₀ = 68 ft/s

∴ Vertical component of velocity = v_y = v₀×sin(θ₀)×t - g×t

Horizontal component of velocity = vₓ = 68×cos(34) = 56.375 ft/s

The equation for projectile motion is as follows

x = x₀ + v₀×cos(θ₀)×t

y = y₀ + v₀×sin(θ₀)×t - 1/2×g×t²

Where:

x₀ = Initial horizontal displacement of the ball = 0

y₀ = Initial vertical displacement of the ball = 0

t = Time of flight of the ball

Therefore;

108 = 0 + 68×cos(34)×t which gives;

t = 108/56.375 = 1.916 seconds

Hence, y = y₀ + v₀×sin(θ₀)×t - 1/2×g×t² gives;

y = 0 + 68×sin(34)×1.916 - 1/2×32.2×1.916²

y = 72.85 - 59.09 = 13.76 ft, hence the vertical location of the ball at 36 yard is 13.76 ft where the crossbar is at 10 ft hence the ball clears the crossbar by 13.76 ft - 10 ft = 3.76 ft ≈ 4 ft

Therefore, the kick is good! The ball clears the crossbar by about 4 feet.

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9x^2y^2

Step-by-step explanation:

Ok so to find the area of a square we have to square the side length. So 3xy*3xy = 9x^2y^2 because 3*3=9, x*x=x^2, and y*y=y^2.

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A 10ft by 20ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216ft^
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check the picture below.


so, we know the dimensions of the pool, is a 20x10, so its area is simply 200 ft², and we know the walkway is 216 ft², so the whole thing, including pool and walkway is really 200 + 216 ft².


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Answer: v <= 1


Step-by-step explanation:

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Answer:

(a) x = -2y

(c) 3x - 2y = 0

Step-by-step explanation:

You can tell if an equation is a direct variation equation if it can be written in the format y = kx.

Note that there is no addition and subtraction in this equation.

Let's put these equations in the form y = kx.

(a) x = -2y

  • y = x/-2 → y = -1/2x
  • This is equivalent to multiplying x by -1/2, so this is an example of direct variation.

(b) x + 2y = 12

  • 2y = 12 - x
  • y = 6 - 1/2x
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(c) 3x - 2y = 0

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  • This follows the format of y = kx, so it is an example of direct variation.

(d) 5x² + y = 0

  • y = -5x²
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(e) y = 0.3x + 1.6

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(f) y - 2 = x

  • y = x + 2
  • 2 is being added to x, so it is <u>NOT</u> an example of direct variation.

The following equations are examples of direct variation:

  • x = -2y
  • 3x - 2y = 0
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