Question

An astronaut on Mars tosses a rock straight. The height, h, measured in feet after T seconds, is given by h(T) = -6T^2 + 24T + 6.

1. After how many seconds will the rock be 20 ft above the surface?

2. After how long will it take for the rock to return to the surface?

First one who answers both questions first gets Brainliest; thanks in advance!

Answer 1

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:

$\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\ x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\\\when ax^2+bx+c=0 \\\end{minipage}}$

Given function:

$h(t)=-6t^2+24t+6$

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.

$\implies h(t)=20$

$\implies -6t^2+24t+6=20$

$\implies -6t^2+24t+6-20=20-20$

$\implies -6t^2+24t-14=0$

$\implies -2(3t^2-12t+7)=0$

$\implies 3t^2-12t+7=0$

Solve using the quadratic formula:

$\implies t=\dfrac{-(-12) \pm \sqrt{(-12)^2-4(3)(7)}}{2(3)}$

$\implies t=\dfrac{12 \pm \sqrt{144-84}}{6}$

$\implies t=\dfrac{12 \pm \sqrt{60}}{6}$

$\implies t=\dfrac{12 \pm \sqrt{4 \cdot 15}}{6}$

$\implies t=\dfrac{12 \pm \sqrt{4} \sqrt{15}}{6}$

$\implies t=\dfrac{12 \pm 2 \sqrt{15}}{6}$

$\implies t=\dfrac{6\pm \sqrt{15}}{3}$

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.

$\implies h(t)=0$

$\implies -6t^2+24t+6=0$

$\implies -6(t^2-4t-1)=0$

$\implies t^2-4t-1=0$

Solve using the quadratic formula:

$\implies t=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-1)}}{2(1)}$

$\implies t=\dfrac{4 \pm \sqrt{16+4}}{2}$

$\implies t=\dfrac{4 \pm \sqrt{20}}{2}$

$\implies t=\dfrac{4 \pm \sqrt{4 \cdot 5}}{2}$

$\implies t=\dfrac{4 \pm \sqrt{4} \sqrt{5}}{2}$

$\implies t=\dfrac{4 \pm 2\sqrt{5}}{2}$

$\implies t=2 \pm\sqrt{5}$

As time is positive,  $t = 2 + \sqrt{5}$  only.

Therefore, it will take 4.24 s (2 d.p.) for the rock to return to the surface.