Chapter 9.1, Problem 65PPS

(a)

To determine

To Write:

The height of football after 1 second.

Answer to Problem 65PPS

The height of football after 1 second = 74 feet.

Explanation of Solution

Given Information:

A football is kicked up from ground level at an initial upward velocity of 90 feet per second. The equation $h=-16t^2+90t$ gives the height h of football after t seconds.

Calculation:

Given t =1 ,

Substituting t =1 in the equation $h=-16t^2+90t$ :

  $\begin{array}{l}h=-16t^2+90t\\ $ h=-16{(1)}^2+90(1)\mathrm{..............}[t=1]$h=-16+90=74\end{array}$

So, the heigh of the football after 1 second = 74 feet.

(b)

To determine

To Find:

The time when the ball is 126 feet high.

Answer to Problem 65PPS

The ball will reach a height of 126 feet on the way up at t = 2.625 seconds and on the way down at t = 3 seconds.

Explanation of Solution

Given Information:

A football is kicked up from ground level at an initial upward velocity of 90 feet per second. The equation $h=-16t^2+90t$ gives the height h of football after t seconds.

Calculation:

Given h =126 . Substituting the value in the equation $h=-16t^2+90t$ and solving for t :

  $\begin{array}{l}h=-16t^2+90t\\ 126=-16t^2+90t\mathrm{.....................}[h=126]\\ 126-126=-16t^2+90t-\mathrm{126........}[subtracting"126"each\_side]\\ 0=-2\cdot -\frac{1}{2}(-16t^2+90t-126)\mathrm{...........}[multiply\&divide"-2"on\_right\_hand\_side]\\ 0=-2(8t^2-45t+63)\mathrm{..........................}[simplify]\\ 0=-2(8t^2-24t-21t+63)\\ 0=-2[(8t^2-24t)-(21t-63)]\\ 0=-2[8t(t-3)-21(t-3)]\mathrm{.............................}[distributivity\_over\_subtraction]\\ 0=-2[(8t-21)(t-3)]\mathrm{.....................................}[distributivity\_over\_subtraction]\\ 0=-2(8t-21)(t-3)\end{array}$

Here, we have the equation of the form xy = 0.

So,

  $\begin{array}{l}8t-21=0\\ 8t-21+21=0+\mathrm{21.......................}[add"21"each\_side]\\ 8t=21\\ \frac{8t}{8}=\frac{21}{8}\mathrm{..........................................}[divide"8"each\_side]\\ t=2.625\end{array}$

Or

  $\begin{array}{l}t-3=0\\ t-3+3=0+\mathrm{3...................}[add"3"each\_side]\\ t=3\end{array}$

So, the ball will reach a height of 126 feet on the way up at t = 2.625 seconds and on the way down at t = 3 seconds.

(c)

To determine

To Find:

The time at which the height of the ball 0 feet and what do these points represent in the context of the situation.

Answer to Problem 65PPS

The height of the ball is 0 feet at t = 0 or t = 5.625.

At t = 0 , the height is 0 since the ball is not yet been kicked.

At t = 5.625 , the height is 0 since the ball hits the ground after the kick.

Explanation of Solution

Given Information:

A football is kicked up from ground level at an initial upward velocity of 90 feet per second. The equation $h=-16t^2+90t$ gives the height h of football after t seconds.

Calculation:

Substitute h = 0 in the equation $h=-16t^2+90t$ and then factor it :

  $\begin{array}{l}h=-16t^2+90t\\ 0=-16t^2+90t\mathrm{......................}[h=0]\\ 0=-2t\cdot \frac{1}{-2t}(-16t^2+90t)\mathrm{.......}[divide\&multiply"-2t"on\_right\_hand\_side]\\ 0=-2t(8t-45)\end{array}$

This is of the form xy = 0.

So,

  $\begin{array}{l}-2t=0\\ \frac{-2t}{-2}=\frac{0}{-2}\mathrm{..........................}[divide"-2"each\_side]\\ t=0\\ or\\ 8t-45=0\\ 8t-45+45=0+\mathrm{45.................}[add"45"each\_side]\\ 8t=45\\ \frac{8t}{8}=\frac{45}{8}\mathrm{...................................}[divide"8"each\_side]\\ t=5.625\end{array}$

The height of the ball is 0 feet at t = 0 or t = 5.625.

At t = 0 , the height is 0 since the ball is not yet been kicked.

At t = 5.625 , the height is 0 since the ball hits the ground after the kick.