Chapter 4.8, Problem 49E

To determine

To find:

The height oh the cliff.

Answer to Problem 49E

  $225ft$

Explanation of Solution

Given:

The acceleration due to gravity on earth at any time $t$

  $a\left(t\right)=-32{\text{ft/s}}^{\text{2}}$

Concept used:

Antiderivative :- A function $F\left(x\right)$ is antiderivative of on an interval I if $F\text{'}\left(x\right)=f\left(x\right)$ for all $x$ in interval I

The entire family of antiderivative of a function by adding a constant to a known antiderivative

So, if $F\left(x\right)$ is the antiderivative of $f\left(x\right)$ , then the family of the antiderivatives would be $F\left(x\right)+c$

Calculation:

The acceleration due to gravity on earth at any time $t$

  $a\left(t\right)=-32{\text{ft/s}}^{\text{2}}$

The antiderivative is the velocity

  $v\left(t\right)=-32t+C$

  $\begin{array}{l}v\left(t\right)=0\\ 0=-32t+C\\ C=0\\ v\left(t\right)=-32\text{ft/s}\end{array}$

If stone hits the ground at $120ft/s$

  $\begin{array}{l}-32t=-120\\ t=\frac{120}{32}\\ t=3.75s\end{array}$

The antiderivative of velocity is the position

  $\begin{array}{l}s\left(t\right)=-32\cdot \frac{1}{2}{t}^2+C_2\\ s\left(t\right)=-16t^2+C_2\\ s\left(3.75\right)=0\\ s\left(3.75\right)=-16t^2+C_2\\ 0=-16{\left(3.75\right)}^2+C_2\\ C_2=225\end{array}$

The position function is

  $s\left(t\right)=-32t^2+225$

The position at time $t=0$ is height of the cliff

  $\begin{array}{l}s\left(0\right)=-32{\left(0\right)}^2+225\\ =225ft\end{array}$