Chapter 1.1, Problem 70E

(a)

To determine

To find: The value of g.

Answer to Problem 70E

The value of g is $1.25\text{m}/{\text{s}}^2$

Explanation of Solution

Given information: $\begin{array}{c}t=4\text{s}\\ y=20m\end{array}$

Calculation:

Substitute the values into the equation and solve for g:

  $\begin{array}{c}$ 20=g{(4)}^2$\Rightarrow g=\frac{20}{16}\\ \Rightarrow g=1.25m/s^2\end{array}$

Therefore the required value of $g=1.25\text{m}/{\text{s}}^2$ .

(b)

To determine

To find: The average velocity for the fall.

Answer to Problem 70E

The velocity is 5 meters per second.

Explanation of Solution

Given information: $\begin{array}{c}t=4\text{s}\\ y=20m\end{array}$

Formula used:

Speed on average is defined as:

  $\frac{\text{Δ}s}{\text{Δ}t}$ , where $s$ position and $t$ is time

Calculation:

Consequently, the typical speed is:

  $\begin{array}{c}\frac{\text{Δ}s}{\text{Δ}t}\\ =\frac{20}{4}\\ =5m/s\end{array}$

Therefore, the required velocity is $5m/s$ .

(c)

To determine

To find: The velocity when it hit the bottom.

Answer to Problem 70E

The velocity is $10\text{m}/\text{s}$ .

Explanation of Solution

Given information: $\begin{array}{c}t=4\text{s}\\ y=20m\end{array}$

Calculation:

From part a) $g=1.25$

The equation is $y=1.25t^2$

The average speed of the rock between $\begin{array}{c}t=4\\ t=4+h\end{array}$ can be found to use the slope formula:

  $\begin{array}{c}\frac{\text{Δ}y}{\text{Δ}t}=\frac{1.25{(4+h)}^2-1.25{(4)}^2}{(4+h)-4}\\ =\frac{1.25\left(16+8h+h^2\right)-20}{h}\\ =\frac{10h+1.25h^2}{h}\\ =10+1.25\text{h}\end{array}$

The speed of the rock is $\underset{h\to 0}{\lim }\frac{\text{Δ}y}{\text{Δ}t}$

The speed of the rock when it hit in the bottom is:

  $\begin{array}{c}\underset{h\to 0}{\lim }(10+1.25h)=10+1.25(0)\\ =10m/s\end{array}$

Therefore the velocity is $10m/s$ .