Chapter 1.1, Problem 22E

Solution

a)

To find: The time taken by the bomb to fall 180 ft.

The time taken by bomb to fall 180 feet is $3.4$ seconds.

Given Information:

An object in free fall will fall $d$feet in $t$seconds, where $d$ and $t$are related by the algebraic model $d=16t^2$

Concept Used:

Use the formula $d=16t^2$ to find the time taken by the bomb to fall at 180 feet.

Calculation:

Substitute 180 for $d$ into $d=16t^2$ .

  $\begin{array}{c}180=16t^2\\ \frac{180}{16}=t^2\\ t=\pm \sqrt{\frac{45}{4}}\\ t\approx \pm 3.4\text{ s}\end{array}$

Since time cannot be negative thus, $t\approx 3.4\text{ s}$ .

Therefore, it will take $3.4$ seconds for the bomb to fall 180 feet.

b)

To find: How high was the airplane when the smoke bomb was dropped.

The airplane is at the height of 2500 feet.

Given Information:

An object in free fall will fall $d$feet in $t$seconds, where $d$ and $t$are related by the algebraic model $d=16t^2$ .

The smoke bomb is in free fall for $12.5$ seconds after it is dropped

Concept Used:

Use the formula $d=16t^2$ to find the height of the airplane when the smoke bomb was dropped.

Calculation:

Substitute $12.5$ for $t$ into $d=16t^2$ .

  $\begin{array}{c}d=16{\left(12.5\right)}^2\\ d=16\left(156.25\right)\\ d=2500\text{ ft}\end{array}$

Therefore, the airplane is at 2500 feet.