Chapter 1.5, Problem 9P

Solution

To find: How long is the ball in the air before your friend catches it.

Friend catch the ball in $1.25$ seconds.

Given information:

You are dropping a ball from a window $29$ feet above the ground to your friend who will catch it $4$ feet above the ground.

Calculation:

Since your friend stand on the ground and catch the ball $4$ feet above the ground, so the height is $\left(29-4=25\right)$ feet.

Using a table:

Write an equation that models the situation using the height function:

  $h=-16t^2+h_0$

Substitute $h=0$ and $h_0=25$ in the above equation:

  $0=-16t^2+25$

Make a table of values for the function. Set the table so that the value of $t$ start at $0$ and increase in increments of $0.1$ .

Through the table to find the time $t$ at which the height $h$ is $0$ feet.

$t$	$h=-16t^2+25$
$0$	$25$
$0.1$	$24.84$
$0.2$	$24.36$
$0.3$	$23.56$
$0.4$	$22.44$
$0.5$	$21$
$0.6$	$19.24$
$0.7$	$17.16$
$0.8$	$14.76$
$0.9$	$12.04$
$1.0$	$9$
$1.1$	$5.64$
$1.2$	$1.96$
$1.3$	$-2.04$
$1.4$	$-6.36$
$1.5$	$-11$
$1.6$	$-15.96$

The table shows that $h=0$ between $t=1.2$ and $t=1.3$ because $h$ has a change of sign.

Therefore, Your friend catch the ball between $1.2$ and $1.3$ seconds after it is dropped.

Using a graph:

Write an equation that models the situation using the height function:

  $h=-16t^2+h_0$

Substitute $h=0$ and $h_0=25$ in the above equation:

  $0=-16t^2+25$

Enter the function $h=-16t^2+25$ into a graphing calculator.

  

Graph the height function. The positive $t-$ value for which $h=0$ using the zero feature. The graph shows that $h=0$ when $t=1.25$ .

Therefore, Friend catch the ball is equal to $1.25$ seconds.