Question

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∘. The field goal posts are 3 meters high.Does he make the field goal?What is the ball's velocity and direction of motion just as it reaches the field goal post

Answer 1

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.