Chapter 1, Problem 6TP

Solution

To find: how long is the horseshoe in the air.

The horseshoe in the air is equal to $1.9$ seconds.

Given information:

The graph shows the height $h$ (in feet) after $t$ seconds of a horseshoe tossed during a game of horseshoes. The initial vertical velocity of the horseshoe is $30$ feet per second.

  

Calculation:

The general equation of the quadratic equation is

  $h\left(t\right)=a{t}^2+bt+c\mathrm{...}\left(1\right)$

On differentiating with respect to $t$ :

  $\begin{array}{c}\frac{dh}{dt}=2at+b\\ v=2at+b\mathrm{...}\left(2\right)\end{array}$

Since the initial vertical velocity of the horseshoe is $30$ feet per second, so substitute $t=0,v=30$ in the equation (2):

  $\begin{array}{c}30=2a\left(0\right)+b\\ 30=0+b\\ b=30\end{array}$

Substitute $b=30$ in the equation (1):

  $h\left(t\right)=a{t}^2+30t+c\mathrm{...}\left(3\right)$

At $t=0,h=2$ , substitute $t=0,h=2$ in the equation (3):

  $\begin{array}{c}2=a{\left(0\right)}^2+30\left(0\right)+c\\ 2=0+0+c\\ c=2\end{array}$

Substitute $c=2$ in the above equation:

  $h\left(t\right)=a{t}^2+30t+2\mathrm{...}\left(4\right)$

Substitute $t=1.95,h=0$ in the equation (4):

  $\begin{array}{c}0=a{\left(1.95\right)}^2+30\left(1.95\right)+2\\ 0=3.8025a+58.5+2\\ 3.8025a+60.5=0\\ a=-15.91\end{array}$

Therefore, the equation of the height is below:

  $h\left(t\right)=-15.91t^2+30t+2$

Now, calculate the vertex points.

  $\begin{array}{c}x=-\frac{b}{2a}\\ x=-\frac{30}{2\left(-15.91\right)}\\ x\approx 0.943\end{array}$

So, the horseshoe in the air is below:

  $\begin{array}{c}2t=2\times 0.943\\ =1.886\\ \approx 1.9\end{array}$

Hence, the horseshoe in the air is $1.9$ seconds.