Chapter 4, Problem 14PT

(a)

To determine

To find: The time to reach the ball $130$ feet by solving equation $h\left(t\right)=-16t^2+112t+6$ .

Answer to Problem 14PT

The ball reaches a height of $130$ feet at $\approx 1.37$ second.

Explanation of Solution

Given information:

The given equation is $h\left(t\right)=-16t^2+112t+6$ ..

Parker throws the ball with an initial of velocity of $112$ feet per second.

Calculation:

Calculate the equation for the height of the ball,

  $\begin{array}{c}h\left(t\right)=-16t^2+112t+6\\ -130=16t^2-112t-6\\ 0=16t^2-112t+124\\ 16t^2-112t+124=0\end{array}$ .

The values are

  $a=16,b=-112,c=124$

  $\begin{array}{c}t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{112\pm \sqrt{{\left(-112\right)}^2+-4\left(-16\right)\left(124\right)}}{2\left(16\right)}\\ =\frac{112\pm \sqrt{12544-7936}}{32}\\ =1.37,5.621\end{array}$

Thus,

  $t\approx 1.37$

Therefore, the ball reaches a height of $130$ feet at $\approx 1.37$ second.

(b)

To determine

To find: When the ball will reach $250$ feet.

Answer to Problem 14PT

The ball will never reach $250$ feet.

Explanation of Solution

Given information:

The given equation is $h\left(t\right)=-16t^2+112t+6$ .

Parker throws the ball with an initial of velocity of $112$ feet per second.

Calculation:

Calculate the equation for the height of the ball,

  $\begin{array}{c}h\left(t\right)=-16t^2+112t+6\\ 16t^2-112t-6=250\\ 16t^2-112t+244=0\end{array}$

Thus,

  $\begin{array}{c}{\left(-112\right)}^2-4\left(16\right)\left(244\right)=12544-15616\\ =-3072\end{array}$

To solve using the discrimination method, it is found that there are no real roots.

Therefore, the ball will never reach $250$ feet.

(c)

To determine

To find: The time to the ball hit the ground.

Answer to Problem 14PT

The ball hit the ground at $7.05$ second.

Explanation of Solution

Given information:

The given equation is $h\left(t\right)=-16t^2+112t+6$ ..

Parker throws the ball with an initial of velocity of $112$ feet per second.

Calculation:

Calculate the equation for the height of the ball,

  $\begin{array}{c}-16t^2+112t+6=0\\ 16t^2-112t-6=0\end{array}$

The values are

  $a=16,b=-112,c=-6$ .

calculate the value of $t$ .

  $\begin{array}{c}t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{112\pm \sqrt{{\left(-112\right)}^2+-4\left(16\right)\left(-6\right)}}{2\left(16\right)}\\ =\frac{112\pm 1\sqrt{12544+384}}{32}\\ =\frac{112\pm 113.701}{32}\end{array}$

Two equations are,

  $t=\frac{112-113.701}{32}=-\frac{1.701}{32}=-0.0531$ .

  $t=\frac{112+113.701}{32}=\frac{225.701}{32}=7.05$ .

Time cannot be negative,

Thus,

  $t=7.05$

Therefore, the ball hit the ground at $7.05$ second.