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A player throws a bean bag from a height of 3 feet with an initial velocity of 34 feet per second at an angle of [tex]$30^{\circ}$[/tex].

1. The parametric equation that represents this situation is:
A. [tex]x = 34 \cos(30^{\circ}) t[/tex] and [tex]y = -16 t^2 + 34 \sin(30^{\circ}) t[/tex]
B. [tex]x = 34 \cos(30^{\circ}) t[/tex] and [tex]y = -16 t^2 + 34 \sin(30^{\circ}) t + 3[/tex]
C. [tex]x = 34 \cos(30^{\circ}) t + 3[/tex] and [tex]y = -16 t^2 + 34 \sin(30^{\circ}) t + 3[/tex]

2. The bean bag's path is part of a curve that is a [tex]\square[/tex].

3. To the nearest second, the bean bag was in the air for [tex]\square \checkmark[/tex] second(s).

Sagot :

GinnyAnsweridn
To analyze the situation where a player throws a bean bag from a height of 3 feet with an initial velocity of 34 feet per second at an angle of [tex]\(30^\circ\)[/tex], we begin by setting up the parametric equations that describe the motion of the bean bag.

The equations are based on the initial height, initial velocity, and the angle of projection. These can be represented as:

[tex]\[ x = (34 \cos(30^\circ)) \, t \][/tex]
[tex]\[ y = -16 t^2 + (34 \sin(30^\circ)) \, t + 3 \][/tex]

Here’s a breakdown of the equations:
- [tex]\( x \)[/tex] represents the horizontal position of the bean bag, which depends on the horizontal component of the initial velocity and time.
- [tex]\( y \)[/tex] represents the vertical position of the bean bag, which accounts for the influence of gravity (expressed as [tex]\(-16 t^2\)[/tex], since the gravitational acceleration is [tex]\(-32 \text{ ft/s}^2\)[/tex] and this term is halved), the vertical component of the initial velocity, and the initial height of 3 feet.

Therefore, the correct parametric equations are given by Option B:
[tex]\[ B. \, x = (34 \cos(30^\circ)) \, t \text{ and } y = -16 t^2 + (34 \sin(30^\circ)) \, t + 3. \][/tex]

Next, we recognize the nature of the path taken by the bean bag. The motion of the bean bag follows a parabolic trajectory because it is influenced by the initial velocity, angle of projection, and the constant acceleration due to gravity.

So, the bean bag's path is part of a curve that is a parabola.

Lastly, to determine the time the bean bag was in the air, we need to solve for [tex]\( t \)[/tex] when the bean bag hits the ground (i.e., [tex]\( y = 0 \)[/tex]). The equation is:

[tex]\[ 0 = -16 t^2 + (34 \sin(30^\circ)) \, t + 3. \][/tex]

Using the details provided, we have:
- The vertical component of the velocity, [tex]\( 34 \sin(30^\circ) \approx 17 \text{ ft/s} \)[/tex].
- The discriminant value in the quadratic formula: [tex]\(\Delta = 480.9999999999999\)[/tex].
- The roots of the quadratic equation: [tex]\( t_1 \approx -0.154116 \)[/tex] and [tex]\( t_2 \approx 1.216616 \)[/tex].

We discard the negative root since time cannot be negative, so the bean bag was in the air for approximately 1.216616 seconds.

To the nearest second, the bean bag was in the air for [tex]\( \boxed{1} \)[/tex] second.
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