Chapter 9.1, Problem 61E

a.

To determine

To Find: The time taken by an earring to hit the ground when it hits from the roof of the building.

  

Answer to Problem 61E

The required time taken by an earring to hit the ground is $\frac{\sqrt{55}}{4}$ or 1.85 seconds.

Explanation of Solution

Given information:

The time t(in seconds) it takes an object to hit the ground is given by $t=\sqrt{\frac{h}{16}}$ , where h is the height in feet from which the object is dropped.

  

Calculation:

Since, the height of the building is 55feet,

Substitute 55 for h and solve for t.

  $t=\sqrt{\frac{h}{16}}$

(Write the formula)

  $t=\sqrt{\frac{55}{16}}$

(Replace h by 55)

  $t=\frac{\sqrt{55}}{\sqrt{16}}$

(Quotient property of square roots)

  $t=\frac{\sqrt{55}}{4}$

(Simplify)

The earring takes about 1.85 seconds to hit the ground.

b.

To determine

To Find: The time taken by an earring to hit the ground if it is dropped from two stories (22 feet) below the roof.

Answer to Problem 61E

The required time taken by an earring to hit the ground is $\frac{\sqrt{33}}{4}$ or 1.44 seconds.

Explanation of Solution

Given information:

The time t(in seconds) it takes an object to hit the ground is given by $t=\sqrt{\frac{h}{16}}$ , where h is the height in feet from which the object is dropped.

The earring is dropped from two stories (22 feet) below the roof.

  

Calculation:

As the earring is dropped from two stories (22 feet) below the roof,

So the new height will be:

  $\begin{array}{l}h=55-22\\ h=33\end{array}$

Substitute 33 for h and solve for t

  $\begin{array}{l}t=\sqrt{\frac{h}{16}}\\ t=\sqrt{\frac{33}{16}}\end{array}$

(Replace h by 33)

  $t=\frac{\sqrt{33}}{\sqrt{16}}$

(Quotient property of square roots)

  $t=\frac{\sqrt{33}}{4}$

(Simplify)

The earring takes about $\frac{\sqrt{33}}{4}$ or 1.44 seconds to hit the ground.