Chapter 1.1, Problem 69E

(a)

To determine

To calculate: The balloon’s average velocity during the first 3 sec of fall.

Answer to Problem 69E

The velocity is $14.7$ meters per second.

Explanation of Solution

Given information: Water balloon is dropped $y=4.9t^{2}m$ in t second from a window above the ground.

Calculation:

Speed on average is defined as:

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

Consequently, the average speed is:

  $\begin{array}{c}\frac{\text{Δ}s}{\text{Δ}t}=\frac{s(3)-s(0)}{3-0}-\frac{\left((4.9){(3)}^2\right)-((4.9)(0))}{3}\\ =\frac{44.1}{3}\\ =14.7m/s\end{array}$

Therefore, the average velocity during the first 3 sec of fall is $14.7m/s$ .

(b)

To determine

To calculate: The velocity at the instant $t=3$ .

Answer to Problem 69E

The velocity is $29.4$ meters per second.

Explanation of Solution

Given information: Water balloon is dropped $y=4.9t^{2}m$ in t second from a window above the ground.

Calculation:

To use the diagram on the left, determine the object's average speed from time $t=3$ to any somewhat later time $t=3+h$ .

  $\begin{array}{}$ y(3)=4.9{(3)}^2$$ y(3+h)=4.9{(3+h)}^2$\end{array}$

Then,

  $\begin{array}{c}\frac{\text{Δ}y}{\text{Δ}t}=\frac{y(3+h)-y(3)}{h}\\ =\frac{4.9{(3+h)}^2-4.9{(3)}^2}{h}\end{array}$

Increase the numerator:

  $\begin{array}{c}\frac{\text{Δ}y}{\text{Δ}t}=\frac{4.9\left(9+6h+h^2\right)-44.1}{h}\\ =\frac{29.4h+4.9h^2}{h}\\ =\frac{h(29.4+4.9h)}{h}\end{array}$

Remove the $h$ ,

  $29.4+4.9h$

Then find the limit as $h$ goes to 0 of $\frac{\text{Δ}y}{\text{Δ}t}$

  $\underset{h\to 0}{\lim }\frac{\text{Δ}y}{\text{Δ}t}=\underset{h\to 0}{\lim }(29.4+4.9h)$

By substitute $h=0$ , calculate the limit.

  $\begin{array}{c}{y}^{\prime }=29.4+4.9(0)\\ =29.4\end{array}$

Therefore, the required velocity is $29.4m/s$ .