Chapter 1.5, Problem 38PS

Solution

To write: an equation giving diver’s height h (in feet) above water after t seconds. Also find how long is the diver in air.

  $h\left(t\right)=-16t^2+40$

  $t=\frac{\sqrt{10}}{2}\approx 1.6\text{ sec}$

Given:

Cliff diver dives off a cliff 40 feet above water.

Property Used:

Height function:

When an object is dropped, its height h (in feet) above the ground after t seconds can be modeled after by the function

  $h\left(t\right)=-16t^2+h_0$

Where $h_0$ is the object’s initial height (in feet).

  $\begin{array}{c}\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}\\ \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\end{array}$

Calculation:

By the height function, the height of the diver h (in feet) above the ground after t seconds is given by:

  $h\left(t\right)=-16t^2+h_0$

Now, initially he was standing at the cliff which is 40 feet above water.

So, $h_0=40\text{ feet}$ .

Thus, the equation giving diver’s height h (in feet) above water after t seconds is:

  $h\left(t\right)=-16t^2+40$ .

Now, when the diver reaches the water, its height becomes 0. So, target is to determine t for which $h\left(t\right)=0$ .

  $0=-16t^2+40$ .

This is equivalent to,

  $16t^2=40$ .

Divide both sides by 16.

  $t^2=\frac{40}{16}$

Take square root of both sides,

  $\begin{array}{c}t=\pm \sqrt{\frac{40}{16}}\\ =\pm \frac{\sqrt{40}}{\sqrt{16}}\\ =\pm \frac{\sqrt{4\cdot 10}}{4}\\ =\pm \frac{\sqrt{4}\cdot \sqrt{10}}{4}\\ =\pm \frac{2\sqrt{10}}{4}\\ =\pm \frac{\sqrt{10}}{2}\end{array}$

Since, t represents time, it is a non negative quantity. Thus, the solution is $t=\frac{\sqrt{10}}{2}\approx 1.6\text{ sec}$ .