To solve the problem, let's break it down step-by-step:
1. Convert the Angle to Radians:
[tex]\[
\text{Angle in radians} = 60^{\circ} \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians}
\][/tex]
2. Set Up the Parametric Equations:
The parametric equations given are:
[tex]\[
x(t) = (30 \cos(60^{\circ})) t
\][/tex]
[tex]\[
y(t) = -9.8 t^2 + (30 \sin(60^{\circ})) t + 1
\][/tex]
3. Calculate the Time When the Line Hits the Ground:
This happens when [tex]\( y(t) = 0 \)[/tex]:
[tex]\[
-9.8 t^2 + (30 \sin(60^{\circ})) t + 1 = 0
\][/tex]
Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where:
[tex]\[
a = -9.8, \quad b = 30 \sin(60^{\circ}), \quad c = 1
\][/tex]
The discriminant is:
[tex]\[
b^2 - 4ac = (30 \sin(60^{\circ}))^2 - 4 \cdot (-9.8) \cdot 1
\][/tex]
4. Solve for the Positive Root to Find the Time [tex]\( t_{\text{ground}} \)[/tex]:
The positive value of [tex]\( t \)[/tex] when [tex]\( y(t) = 0 \)[/tex] gives us the time at which the line hits the ground. After solving, we find:
[tex]\[
t_{\text{ground}} = 5.2 \text{ seconds}
\][/tex]
5. Calculate the Horizontal Distance:
Using the horizontal parametric equation:
[tex]\[
x(t) = (30 \cos(60^{\circ})) t_{\text{ground}}
\][/tex]
Plug in the values to find the horizontal distance:
[tex]\[
x(t_{\text{ground}}) = (30 \cos(60^{\circ})) \times 5.2 \approx 40 \text{ meters}
\][/tex]
6. Find the Time to Reach Maximum Height:
The time to reach maximum height is when the vertical velocity component reaches 0. This is given by:
[tex]\[
t_{\text{max\_height}} = \frac{30 \sin(60^{\circ})}{9.8} \approx 2.7 \text{ seconds}
\][/tex]
Thus, based on these calculations:
- The line travels a horizontal distance of approximately 40 meters.
- The end of the line reaches its maximum height after approximately 2.7 seconds.
So, the completed statements will be:
To the nearest meter, the line travels a horizontal distance of 40 meters.
To the nearest tenth of a second, the end of the line reaches its maximum height after 2.7 seconds.
