Chapter 4, Problem 43RE

To determine

To find : How high does it go?

Answer to Problem 43RE

The answer: $1162.5\text{m}$

Explanation of Solution

Given information:

The acceleration of the gravity near the surfaces of Mars is $3.72\text{m}/{\text{sec}}^2$ .

The surface with an velocity of $93\text{m}/\text{sec}$ .

Calculation:

Given that,

  $a(t)=-3.72\left(\text{m}/{\text{sec}}^2\right)$

When a vector is pointing up from the surface, use a positive sign; when it is pointing down, use a negative sign.

Velocity $v$ is the anti derivative of the acceleration.

So,

  $v(t)=-3.72t+C\mathrm{.....}(1)$

Given, $v(0)=93\text{m}/\text{sec}$ , by put

  $t=0$ into equation $(1)$ and obtain:

  $\begin{array}{c}93=-3.72\cdot 0+C\\ \Rightarrow C=93\\ v(t)=-3.72t+93(\text{m}/\text{sec})\\ \Rightarrow -3.72t+93=0\\ \Rightarrow t=25\end{array}$

Since, $v(t)=-3.72t+93$ and get:

  $s(t)=-1.86t^2+93t+C\mathrm{.....}(2)$

A rock is blasted from the surface $\Rightarrow s(0)=0$

Put, $t=0$ into equation $(2)$ and get:

  $\begin{array}{c}0=-1.86\cdot 0+93\cdot 0+C\\ \Rightarrow C=0\end{array}$

So,

  $s(t)=-1.86t^2+93t$

At,

  $\begin{array}{c}t=25\\ (v(25)=0\end{array}$

Then,

  $\begin{array}{c}s(25)=-1.86\cdot {25}^2+93\cdot 25\\ =1162.5\text{m}\end{array}$

Therefore, the required $1162.5\text{m}$ high does it go.