Chapter 4.2, Problem 43E

(a)

To determine

To find: how fast it will be going for it hits bottom $30\sec$ later.

Answer to Problem 43E

  $48\text{m}/\text{s}$ fast it will be going for it hits bottom $30\sec$ later.

Explanation of Solution

Given information:

The acceleration due to gravity is $1.6\text{m}/{\text{sec}}^2$ .

Calculation:

  $a(t)=1.6\text{m}/{\text{sec}}^2$

Velocity is anti derivative of acceleration:

  $v(t)=1.6t+C$

Hence the rock is dropped from rest, initial velocity at $(t=0)$ is $0$ , so then

  $C=0$ get,

  $v(t)=1.6t$

Put, $t=30$ :

  $\begin{array}{c}v(30)=1.6(30)\\ =48\text{m}/\text{s}\end{array}$

Therefore the required $48\text{m}/\text{s}$ fast it will be going for it hits bottom $30\sec$ later.

(b)

To determine

To find: How far below the point is release is the bottom of the crevasse.

Answer to Problem 43E

  $720\text{m}$ far below the point is release is the bottom of the crevasse.

Explanation of Solution

Given information:

The acceleration due to gravity is $1.6\text{m}/{\text{sec}}^2$ .

Calculation:

The position function is the anti derivative of $v(t)$ .

The position at $t=0$ is $0$ , so

  $C=0$

Put, $t=30$ :

  $\begin{array}{c}s(t)=1.6\left(\frac{1}{2}{t}^2\right)+C\\ =0.8t^2\\ $ s(30)=0.8{(30)}^2$=720\text{m}\end{array}$

Therefore the required $720\text{m}$ far below the point is release is the bottom of the crevasse.

(c)

To determine

To find: how fast it will be going and when it does.

Answer to Problem 43E

It will hit in $27.60\text{s}$ is going $48.16\text{m}/\text{s}$

Explanation of Solution

Given information:

The acceleration due to gravity is $1.6\text{m}/{\text{sec}}^2$ .

Calculation:

When $t=0$ is $\text{4m}/\text{s}$ set $v(0)=4$ and solve for $C$ :

  $\begin{array}{c}v(0)=4\\ =1.6(0)+C\\ C=4\end{array}$

  $v(t)=1.6t+4$

Solve for $t$ :

  $\begin{array}{c}s(t)=0.8t^2+4t+C\\ =0.8t^2+4t\\ 720=0.8t^2+4t\\ \Rightarrow 0=0.8t^2+4t-720\\ =t^2+5t-900\\ \Rightarrow t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-(5)\pm \sqrt{{(5)}^2-4(1)(-900)}}{2(1)}\\ =\frac{-5\pm \sqrt{3625}}{2}\\ =\frac{-5+5\sqrt{145}}{2}\\ \approx 27.60\\ =\frac{-5-5\sqrt{145}}{2}\\ \approx -32.60\end{array}$

Take the positive time value and plug it into the velocity equation.

  $\begin{array}{c}v(27.60)=1.6(27.60)+4\\ =48.16\text{m}/\text{s}\end{array}$

Therefore the required It will hit in $27.60\text{s}$ is going $48.16\text{m}/\text{s}$