Chapter 4.2, Problem 37E

(a)

To determine

To Find: The graph that relates the woman’s distance travelled with the time since the jump.

Answer to Problem 37E

The graph is shown in Figure 1.

Explanation of Solution

Given:

The given function is $d\left(t\right)=v_0\left(t\right)-\frac{1}{2}g{t}^2$

The upward initial velocity is $5\text{feet}$ per second.

Calculation:

The given function is,

  $d\left(t\right)=v_0\left(t\right)-\frac{1}{2}g{t}^2$

Then,

  $\begin{array}{c}d\left(t\right)=5\left(t\right)-\frac{1}{2}\left(32\text{ft}{\text{sec}}^{-2}\right)t^2\\ =5t-16t^2\end{array}$

The graph for the above function is shown in Figure 1

  

Figure 1

(b)

To determine

To Find: Thex intercept of the graph.

Answer to Problem 37E

The x intercept of the graph is 0 and 0.3.

Explanation of Solution

Consider the graph shown in Figure 1

From the graph the x intercept of the graph is 0 and 0.3.

(c)

To determine

To Find: The relevance of the x intercept of the graph.

Answer to Problem 37E

The stunt woman is at the same height as the beginning.

Explanation of Solution

The $x$ intercept of the graph is 0 and 0.3 and this shows that the stunt woman is at the same height as the beginning.

(d)

To determine

To Find: The equation that is used to determine the time when the stunt woman reaches the safety pad on the ground.

Answer to Problem 37E

The required function is $5t-16t^2-50=0$ .

Explanation of Solution

Given:

The ground is 50 feet from the starting point.

Calculation:

The given function is,

  $d\left(t\right)=v_0\left(t\right)-\frac{1}{2}g{t}^2$

Then,

  $\begin{array}{l}50=5t-16t^2\\ 5t-16t^2-50=0\end{array}$

(e)

To determine

To Find: The time taken by the stunt woman to reach the safety pad on the ground.

Answer to Problem 37E

The required function is $1.93\sec$ .

Explanation of Solution

Consider the required equation is,

  $5t-16t^2-50=0$

Then,

  $\begin{array}{l}t=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)-4\left(16\right)\left(-50\right)}}{2\left(16\right)}\\ t=\frac{5\pm \sqrt{3225}}{32}\\ t=1.93,-1.62\end{array}$