Question

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

$H\left(x\right)=20-4.9x^2$

(a) What is the height of the rock when $x=1$second? $x=1.1$seconds? $x=1.2$seconds? $x=1.3$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 $H\left(1\right)=15.1$m, when $x=1.1$is role="math" localid="1646327053446" $H(1.1)=14.071$m, when $x=1.2$is $H(1.2)=12.944$m, when $x=1.3$is $H(1.3)=11.719$m

(b) The height of the rock $15$m in $x=1.01$seconds, $10$m in $x=1.43$seconds, $5$m in $x=1.75$seconds.

(c) The rock strike the ground in$x=2.02$seconds.

Step-by-step solution

Step 1. Given Information

If a rock falls from a height of 20 meters on Earth, the height H (in meters) after x seconds is approximately $H\left(x\right)=20-4.9x^2$

(a) What is the height of the rock when $x=1$second? $x=1.1$seconds? $x=1.2$seconds? $x=1.3$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)=20-4.9x2We have to find the height of the rock when x=1 second.

Putting the value of xin the given function.

$$\begin{gathered} H\left(1\right)=20-4.9{\left(1\right)}^2 \\ H\left(1\right)=20-4.9\times 1 \\ H\left(1\right)=20-4.9 \\ H\left(1\right)=15.1 \end{gathered}$$

Part (a) Step 2. We have to find the height of the rock when x=1.1  second.

Putting the value of xin the given function.

$$\begin{gathered} H(1.1)=20-4.9{(1.1)}^2 \\ H(1.1)=20-4.9\times 1.21 \\ H(1.1)=20-5.929 \\ H(1.1)=14.071 \end{gathered}$$

Part (a) Step 3. We have to find the height of the rock when x=1.2  second.

Putting the value of xin the given function.

$$\begin{gathered} H(1.2)=20-4.9{(1.2)}^2 \\ H(1.2)=20-4.9\times 1.44 \\ H(1.2)=20-7.056 \\ H(1.2)=12.944 \end{gathered}$$

Part (a) Step 4. We have to find the height of the rock whenx=1.3 seconds.

Putting the value of xin the given function.

$$\begin{gathered} H(1.3)=20-4.9{(1.3)}^2 \\ H(1.3)=20-4.9\times 1.69 \\ H(1.3)=20-8.281 \\ H(1.3)=11.719 \end{gathered}$$

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

$15=20-4.9x^2$

Subtract 20 on both side

$$\begin{gathered} 15-20=20-4.9x^2-20 \\ -5=-4.9x^2 \end{gathered}$$

Divide by $-4.9$on both side

localid="1646330439926" $$\begin{gathered} \frac{-5}{-4.9}=\frac{-4.9}{-4.9}{x}^2 \\ 1.02=x^2 \\ {x}^2=1.02 \end{gathered}$$

Taking square root on both side

$$\begin{gathered} x=\sqrt{1.02} \\ x=1.01 \end{gathered}$$

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

$10=20-4.9x^2$

Subtract 20 on both side

localid="1646330693702" $$\begin{gathered} 10-20=20-4.9x^2-20 \\ -10=-4.9x^2 \end{gathered}$$

Divide by on both side

$$\begin{gathered} \frac{-10}{-4.9}=\frac{-4.9}{-4.9}{x}^2 \\ 2.04=x^2 \\ {x}^2=2.04 \end{gathered}$$

Taking square root on both side

$$\begin{gathered} x=\sqrt{2.04} \\ x=1.43 \end{gathered}$$

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

$5=20-4.9x^2$

Subtract 20 on both side

$$\begin{gathered} 5-20=20-4.9x^2-20 \\ -15=-4.9x^2 \end{gathered}$$

Divide by on both side

localid="1646330768641" $$\begin{gathered} \frac{-15}{-4.9}=\frac{-4.9}{-4.9}{x}^2 \\ 3.06=x^2 \\ {x}^2=3.06 \end{gathered}$$

Taking square root on both side

$$\begin{gathered} x=\sqrt{3.06} \\ x=1.75 \end{gathered}$$

Part (c) Step 1. We have to find when does the rock strike the ground.

When the rock strike the ground then $H\left(x\right)=0$

So $0=20-4.9x^2$

Subtract by 20 on both side

$$\begin{gathered} 0-20=20-4.9x^2-20 \\ -20=-4.9x^2 \end{gathered}$$

Divide by $-4.9$on both side

$$\begin{gathered} \frac{-20}{-4.9}=\frac{-4.9}{-4.9}{x}^2 \\ 4.08=x^2 \\ {x}^2=4.08 \end{gathered}$$

Taking square root on both side

$$\begin{gathered} x=\sqrt{4.08} \\ x=2.02 \end{gathered}$$