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2 years ago

Write the equation of a sine or cosine function to describe the graph.

Mathematics

1 answer:

prohojiy [21]2 years ago

5 0

Y = A sin(Bx) and y = A cos(Bx). The number, A, in front of sine or cosine changes the height of the graph. The value A (in front of sin or cos) affects the amplitude (height). The amplitude (half the distance between the maximum and minimum values of the function) will be |A|, since distance is always positive.

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Is number 1 and 2 correct and can someone help me with number 3

IrinaK [193]

H might be "4".

I'm sorry if it is wrong because it is too small.

4 0

3 years ago

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:

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