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koban [17]
2 years ago
11

Bike sprockets and chain.

Mathematics
1 answer:
Delvig [45]2 years ago
5 0

Answer:

4.63

Step-by-step explanation:

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A field goal kicker lines up to kick a 44 yard (40m) field goal. He kicks it with an initial velocity of 22m/s at an angle of 55
Goryan [66]

Answer:

The ball makes the field goal.

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of motion is -45.999º or 314.001º.

Step-by-step explanation:

According to the statement of the problem, we notice that ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical uniform accelerated motion, whose equations of motion are described below:

x = x_{o}+v_{o}\cdot t\cdot \cos \theta (Eq. 1)

y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2}\cdot g\cdot t^{2} (Eq. 2)

Where:

x_{o}, y_{o} - Coordinates of the initial position of the ball, measured in meters.

x, y - Coordinates of the final position of the ball, measured in meters.

\theta - Angle of elevation, measured in sexagesimal degrees.

v_{o} - Initial speed of the ball, measured in meters per square second.

t - Time, measured in seconds.

If we know that x_{o} = 0\,m, y_{o} = 0\,m, v_{o} = 22\,\frac{m}{s}, \theta = 55^{\circ}, g = -9.807\,\frac{m}{s} and x = 40\,m, the following system of equations is constructed:

40 = 12.618\cdot t (Eq. 1b)

y = 18.021\cdot t -4.904\cdot t^{2} (Eq. 2b)

From (Eq. 1b):

t = 3.170\,s

And from (Eq. 2b):

y = 7.847\,m

Therefore, the ball makes the field goal.

In addition, we can calculate the components of the velocity of the ball when it reaches the field goal post by means of these kinematic equations:

v_{x} = v_{o}\cdot \cos \theta (Eq. 3)

v_{y} = v_{o}\cdot \cos \theta + g\cdot t (Eq. 4)

Where:

v_{x} - Final horizontal velocity, measured in meters per second.

v_{y} - Final vertical velocity, measured in meters per second.

If we know that v_{o} = 22\,\frac{m}{s}, \theta = 55^{\circ}, g = -9.807\,\frac{m}{s} and t = 3.170\,s, then the values of the velocity components are:

v_{x} = \left(22\,\frac{m}{s} \right)\cdot \cos 55^{\circ}

v_{x} = 12.619\,\frac{m}{s}

v_{y} = \left(22\,\frac{m}{s} \right)\cdot \sin 55^{\circ} +\left(-9.807\,\frac{m}{s^{2}} \right)\cdot (3.170\,s)

v_{y} = -13.067\,\frac{m}{s}

The magnitude of the final velocity of the ball is determined by Pythagorean Theorem:

v =\sqrt{v_{x}^{2}+v_{y}^{2}} (Eq. 5)

Where v is the magnitude of the final velocity of the ball.

If we know that v_{x} = 12.619\,\frac{m}{s} and v_{y} = -13.067\,\frac{m}{s}, then:

v = \sqrt{\left(12.619\,\frac{m}{s} \right)^{2}+\left(-13.067\,\frac{m}{s}\right)^{2} }

v \approx 18.166\,\frac{m}{s}

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of the final velocity is given by this trigonometrical relation:

\theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}} \right) (Eq. 6)

Where \theta is the angle of the final velocity, measured in sexagesimal degrees.

If we know that v_{x} = 12.619\,\frac{m}{s} and v_{y} = -13.067\,\frac{m}{s}, the direction of the ball is:

\theta = \tan^{-1}\left(\frac{-13.067\,\frac{m}{s} }{12.619\,\frac{m}{s} } \right)

\theta = -45.999^{\circ} = 314.001^{\circ}

The direction of motion is -45.999º or 314.001º.

8 0
3 years ago
A construction company charges $15 per hour for debris removal, plus one time fee for the use of the trash computer. The total f
Svetach [21]

Let h be the numbers of hours, f be the one-time fee and c the cost charged. The equation is

c=15h+f

Since you pay $15 per four, plus the fee. We can solve this equation for the fee:

f=c-15h

If we plug c=195 and h=9, we have

f=195-15\cdot 9=195-135=60

4 0
3 years ago
The given box plots show the number of text messages Paul and Sally received each day on their cell phones.
zavuch27 [327]
To be able to answer this we need to see the box plots and the rest of the question. Sorry -♡
7 0
4 years ago
In your own words what would the steps be to solve by graphing.
expeople1 [14]

Answer:

Graph the first equation.

Graph the second equation on the same rectangular coordinate system.

Determine whether the lines intersect, are parallel, or are the same line.

Identify the solution to the system. If the lines intersect, identify the point of intersection. ...

Check the solution in both equations.

Step-by-step explanation:

7 0
3 years ago
Helpppppp mathhhhhhhhhh
Alex Ar [27]
The final answer will be 14.4
6 0
4 years ago
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