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A coach throws a football from a height of 4 feet with an initial velocity of 59 feet per second at an elevation angle of [tex]$45^{\circ}$[/tex]. Which parametric equations represent the path of the football?

A. [tex]x(t)=59 \sin \left(45^{\circ}\right) t[/tex] and [tex]y(t)=-16 t^2+59 \cos \left(45^{\circ}\right) t[/tex]

B. [tex]x(t)=59 \cos \left(45^{\circ}\right) t[/tex] and [tex]y(t)=-16 t^2+59 \sin \left(45^{\circ}\right) t[/tex]

C. [tex]x(t)=59 \cos \left(45^{\circ}\right) t[/tex] and [tex]y(t)=-16 t^2+59 \sin \left(45^{\circ}\right) t+4[/tex]

D. [tex]x(t)=59 \sin \left(45^{\circ}\right) t[/tex] and [tex]y(t)=-16 t^2+59 \cos \left(45^{\circ}\right) t+4[/tex]

Answer :

To find the parametric equations representing the path of the football, we need to consider both the horizontal and vertical components of the football's motion. Here's a step-by-step breakdown:

1. Initial Conditions:
- Initial height of the throw: 4 feet.
- Initial velocity of the football: 59 feet per second.
- Angle of projection: [tex]\(45^\circ\)[/tex].

2. Understanding Components:
- The initial velocity can be broken down into horizontal and vertical components using trigonometry:
- Horizontal component: [tex]\(v_x = v \cdot \cos(\theta)\)[/tex]
- Vertical component: [tex]\(v_y = v \cdot \sin(\theta)\)[/tex]
- In this case, [tex]\(\theta = 45^\circ\)[/tex].

3. Conversion and Calculations:
- [tex]\(\cos(45^\circ)\)[/tex] and [tex]\(\sin(45^\circ)\)[/tex] both equal [tex]\(\frac{\sqrt{2}}{2}\)[/tex] or approximately 0.7071. However, here we directly keep trigonometric functions due to simplicity.
- [tex]\(v_x = 59 \cdot \cos(45^\circ)\)[/tex]
- [tex]\(v_y = 59 \cdot \sin(45^\circ)\)[/tex]

4. Equations for Motion:
- Horizontal Motion: The horizontal position [tex]\(x(t)\)[/tex] at time [tex]\(t\)[/tex] is given by:
[tex]\[
x(t) = v_x \cdot t = 59 \cdot \cos(45^\circ) \cdot t
\][/tex]
- Vertical Motion: The vertical position [tex]\(y(t)\)[/tex] accounts for initial height and gravity, using:
[tex]\[
y(t) = -16t^2 + v_y \cdot t + \text{initial height}
\][/tex]
[tex]\[
y(t) = -16t^2 + 59 \cdot \sin(45^\circ) \cdot t + 4
\][/tex]
- Note: The [tex]\( -16t^2 \)[/tex] term comes from the gravitational acceleration converted to feet per second squared.

5. Parametric Equations:
- Therefore, the parametric equations that represent the path of the football are:
[tex]\[
x(t) = 59 \cdot \cos(45^\circ) \cdot t
\][/tex]
[tex]\[
y(t) = -16t^2 + 59 \cdot \sin(45^\circ) \cdot t + 4
\][/tex]

From the options provided, the correct choice matches this breakdown:

- [tex]\(x(t) = 59 \cdot \cos(45^\circ) \cdot t\)[/tex]
- [tex]\(y(t) = -16 \cdot t^2 + 59 \cdot \sin(45^\circ) \cdot t + 4\)[/tex]

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Rewritten by : Barada