Q.95 - Precalculus Enhanced with Graphing Utilities

Question

Gravity on Earth If a rock falls from a height of$100$meters

on Earth, the height H (in meters) after t seconds is approximately

$H\left(t\right)=100-4.9t^2$

(a) In general, quadratic functions are not one-to-one. However, the function H is one-to-one. Why?

(b) Find the inverse of H and verify your result.

(c) How long will it take a rock to fall $80$ meters?

Step-by-Step Solution

Answer

Part (a). We know that there are two solutions of every quadratic function. But time is always positive. So, this function is one-to one function.

Part (b). The inverse of T is $t=\sqrt{\frac{100-H\left(t\right)}{4.9}}$. It is verified that $t$ is the inverse of H.

Part (c). It will take $2.02secs$ to fall from the height $80m$.

Part (a) Step 1. Explanation

We know that there are two solutions of every quadratic function. But time is always positive. So, this function is one-to one function.

Part (b) Step 1. Find the inverse

The given equation is $H\left(t\right)=100-4.9t^2$

Finding the inverse, we get:

$$\begin{gathered} H\left(t\right)=100-4.9t^2 \\ H\left(t\right)-100=-4.9t^2 \left[subtracting 100 from each side\right] \\ \frac{H\left(t\right)-100}{-4.9}=t^2 \left[divide 4.9 by each side\right] \\ \frac{100-H\left(t\right)}{4.9}=t^2 \\ t=\sqrt{\frac{100-H\left(t\right)}{4.9}} \end{gathered}$$

Verifying the inverse by substituting $H\left(t\right)$ in $t\left(H\left(t\right)\right)=t$, we get:

$$\begin{gathered} t\left(H\left(t\right)\right)=t \\ t\left(\frac{100-H}{4.9}\right)=t \\ \sqrt{\frac{100-100+4.9t^2}{4.9}}=t \\ \sqrt{\frac{4.9t^2}{4.9}}=t \\ \sqrt{t^2}=t \\ t=t \end{gathered}$$

This is true

Verifying the inverse by substituting $t\left(H\right)$ in $H\left(t\left(H\right)\right)=H$, we get:

$$\begin{gathered} H\left(t\left(H\right)\right)=H \\ 100-4.9{\left(\sqrt{\frac{100-H}{4.9}}\right)}^2=H \\ 100-4.9\left(\frac{100-H}{4.9}\right)=H \\ 100-100+H=H \\ H=H \end{gathered}$$

This is true

Part (c) Step 1. Find the time

Substituting $H=80m$ in the function $t=\sqrt{\frac{100-H\left(t\right)}{4.9}}$ , we get:

$$\begin{gathered} t=\sqrt{\frac{100-H\left(t\right)}{4.9}} \\ t=\sqrt{\frac{100-80}{4.9}} \\ t=\sqrt{\frac{20}{4.9}} \\ t=2.02secs \end{gathered}$$