Answer:
1. 0.71 s (2 d.p) and 3.29 s (2 d.p)
2. 4.24 s (2 d.p.)
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}[/tex]
Given function:
[tex]h(t)=-6t^2+24t+6[/tex]
where:
- h is the height above the surface (in feet).
- t is the time (in seconds).
Question 1
To find the time at which the rock will be 20 ft above the surface, substitute h = 20 into the given function and solve for t.
[tex]\implies h(t)=20[/tex]
[tex]\implies -6t^2+24t+6=20[/tex]
[tex]\implies -6t^2+24t+6-20=20-20[/tex]
[tex]\implies -6t^2+24t-14=0[/tex]
[tex]\implies -2(3t^2-12t+7)=0[/tex]
[tex]\implies 3t^2-12t+7=0[/tex]
Solve using the quadratic formula:
[tex]\implies t=\dfrac{-(-12) \pm \sqrt{(-12)^2-4(3)(7)}}{2(3)}[/tex]
[tex]\implies t=\dfrac{12 \pm \sqrt{144-84}}{6}[/tex]
[tex]\implies t=\dfrac{12 \pm \sqrt{60}}{6}[/tex]
[tex]\implies t=\dfrac{12 \pm \sqrt{4 \cdot 15}}{6}[/tex]
[tex]\implies t=\dfrac{12 \pm \sqrt{4} \sqrt{15}}{6}[/tex]
[tex]\implies t=\dfrac{12 \pm 2 \sqrt{15}}{6}[/tex]
[tex]\implies t=\dfrac{6\pm \sqrt{15}}{3}[/tex]
Therefore, the rock will be 20 ft above the surface at:
- 0.71 s (2 d.p)
- 3.29 s (2 d.p)
Question 2
The rock will return to the surface when h = 0. Therefore, substitute h = 0 into the given function and solve for t.
[tex]\implies h(t)=0[/tex]
[tex]\implies -6t^2+24t+6=0[/tex]
[tex]\implies -6(t^2-4t-1)=0[/tex]
[tex]\implies t^2-4t-1=0[/tex]
Solve using the quadratic formula:
[tex]\implies t=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-1)}}{2(1)}[/tex]
[tex]\implies t=\dfrac{4 \pm \sqrt{16+4}}{2}[/tex]
[tex]\implies t=\dfrac{4 \pm \sqrt{20}}{2}[/tex]
[tex]\implies t=\dfrac{4 \pm \sqrt{4 \cdot 5}}{2}[/tex]
[tex]\implies t=\dfrac{4 \pm \sqrt{4} \sqrt{5}}{2}[/tex]
[tex]\implies t=\dfrac{4 \pm 2\sqrt{5}}{2}[/tex]
[tex]\implies t=2 \pm\sqrt{5}[/tex]
As time is positive, [tex]t = 2 + \sqrt{5}[/tex] only.
Therefore, it will take 4.24 s (2 d.p.) for the rock to return to the surface.
