Chapter 2.4, Problem 27E

To determine

To find: The speed of the fall after $2\sec$ of drop.

Answer to Problem 27E

The speed of the fall after $2\sec$ of drop is equal to $19.6\text{m}/\sec$.

Explanation of Solution

Given information: An object is dropped from the top of $100\text{m}$ tower. Its height above the ground after $t\sec$ is $100-4.9t^2\text{m}$.

Calculation:

The speed of the fall of ball after $t\sec$ is the instantaneous rate of change of height function with respect to time.

The formula for the instantaneous rate of change of height with respect to time at $t$ seconds is equal to $\underset{h\to 0}{\lim }\frac{f\left(t+h\right)-f\left(t\right)}{h}$.

The formula for the instantaneous rate of change of $y=100-4.9t^2\text{m}$ at $t=2\sec$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$.

The function is given as $f\left(t\right)=100-4.9t^2$.

Substitute $2+h$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=100-4.9t^2\\ f\left(t\right)=100-4.9{\left(2+h\right)}^2\\ f\left(t\right)=100-4.9\left(4+h^2+4h\right)\\ f\left(t\right)=80.4-4.9h^2-19.6h\end{array}$

Substitute $2$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=100-4.9t^2\\ f\left(2\right)=100-4.9{\left(2\right)}^2\\ f\left(2\right)=100-19.6\\ f\left(2\right)=80.4\end{array}$

Substitute $80.4-4.9h^2-19.6h$ for $f\left(2+h\right)$ and $80.4$ for $f\left(2\right)$ in the formula.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\frac{80.4-4.9h^2-19.6h-80.4}{h}\\ =\underset{h\to 0}{\lim }\frac{-4.9h^2-19.6h}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{-h\left(4.9h+19.6\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(-\left(4.9h+19.6\right)\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=-\left(4.9\left(0\right)+19.6\right)\\ =-19.6\end{array}$

Negative sign of the speed shows that the height is decreasing with respect to time.

Therefore, the speed of the fall after $2\sec$ of drop is equal to $19.6\text{m}/\sec$.