Question

Effect of Gravity on Jupiter If a rock falls from a height of 20 meters on the planet Jupiter, its height H (in meters) after x seconds is approximately

$H\left(x\right)=20-13x^2$

(a) What is the height of the rock when $x=1$second? $x=1.1$seconds?$x=1.2$ seconds?

(b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters?

(c) When does the rock strike the ground?

Short answer

(a) The height of the rock when $x=1$is $H\left(1\right)=7$m, when $x=1.1$is $H(1.1)=4.27$m, when $x=1.2$is $H(1.2)=1.28$m.

(b) The height of the rock $15$m in $x\approx 0.62$seconds, $10$m in $x\approx 0.88$seconds, $5$m in $x\approx 1.08$seconds.

(c) The rock strike the ground in $x=1.24$seconds.

Step-by-step solution

Step 1. Given Information

Effect of Gravity on Jupiter If a rock falls from a height of 20 meters on the planet Jupiter, its height H (in meters) after x seconds is approximately

$H\left(x\right)=20-13x^2$

(a) What is the height of the rock when $x=1$second? $x=1.1$seconds? $x=1.2$seconds?

(b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters?

(c) When does the rock strike the ground?

Part (a) Step 1. The given function is H(x)=20-13x2We have to find the height of the rock when x=1 second.

Putting the value of xin the given function.

$$\begin{gathered} H\left(1\right)=20-13{\left(1\right)}^{2}H\left(1\right)=20-13\times 1 \\ H\left(1\right)=20-13 \\ H\left(1\right)=7 \end{gathered}$$

Part (a) Step 2. We have to find the height of the rock when   second.

Putting the value of xin the given function.

$$\begin{gathered} H(1.1)=20-13{(1.1)}^2 \\ H(1.1)=20-13\times 1.21 \\ H(1.1)=20-15.73 \\ H(1.1)=4.27 \end{gathered}$$

Part (a) Step 3. We have to find the height of the rock when   second. 

Putting the value of xin the given function.

$$\begin{gathered} H(1.2)=20-13{(1.2)}^2 \\ H(1.2)=20-13\times 1.44 \\ H(1.2)=20-18.72 \\ H(1.2)=1.28 \end{gathered}$$

Part (b) Step 1. We have to find the seconds x when the height of the rock 15 meters. 

Putting $H\left(x\right)=15$in given function

$15=20-13x^2$

Subtract 20 on both side

$$\begin{gathered} 15-20=20-13x^2-20 \\ -5=-13x^2 \end{gathered}$$

Divide by $-13$on both side

$$\begin{gathered} \frac{-5}{-13}=\frac{-13}{-13}{x}^2 \\ 0.385\approx x^2 \\ {x}^2\approx 0.385 \end{gathered}$$

Taking square root on both side

$$\begin{gathered} x\approx \sqrt{0.385} \\ x\approx 0.62 \end{gathered}$$

Part (b) Step 2. We have to find the seconds x when the height of the rock 10 meters. 

Putting $H\left(x\right)=10$in given function

localid="1646398133825" $10=20-13x^2$

Subtract 20 on both side

$$\begin{gathered} 10-20=20-13x^2-20 \\ -10=-13x^2 \end{gathered}$$

Divide by $-13$on both side

localid="1646366229509" $$\begin{gathered} \frac{-10}{-13}=\frac{-13}{-13}{x}^2 \\ 0.769\approx x^2 \\ {x}^2\approx 0.769 \end{gathered}$$

Taking square root on both side

localid="1646366265870" $$\begin{gathered} x\approx \sqrt{0.769} \\ x\approx 0.88 \end{gathered}$$

Part (b) Step 3. We have to find the seconds x when the height of the rock 5 meters. 

Putting $H\left(x\right)=5$in given function

$5=20-13x^2$

Subtract 20 on both side

$$\begin{gathered} 5-20=20-13x^2-20 \\ -15=-13x^2 \end{gathered}$$

Divide by $-13$on both side

localid="1646366329071" $$\begin{gathered} \frac{-15}{-13}=\frac{-13}{-13}{x}^2 \\ 1.154\approx x^2 \\ {x}^2\approx 1.154 \end{gathered}$$

Taking square root on both side

localid="1646366357100" $$\begin{gathered} x\approx \sqrt{1.154} \\ x\approx 1.08 \end{gathered}$$

Part (c) Step 1. We have to find when does the rock strike the ground.When the rock strike the ground then H(x)=0

Putting $H\left(x\right)=0$in given function

$0=20-13x^2$

Subtract 20 on both side

$$\begin{gathered} 0-20=20-13x^2-20 \\ -20=-13x^2 \end{gathered}$$

Divide by $-13$on both side

$$\begin{gathered} \frac{-20}{-13}=\frac{-13}{-13}{x}^2 \\ 1.538=x^2 \\ {x}^2=1.538 \end{gathered}$$

Taking square root on both side

$$\begin{gathered} x=\sqrt{1.538} \\ x=1.24 \end{gathered}$$