Chapter 10, Problem 83RE

a.

To determine

Find parametric equations that model the position of the ball as a function of time.

Answer to Problem 83RE

  $\boxed{3\sec onds}$

Explanation of Solution

Given information:

Drew Brees throws a football with an initial speed of $80feet/\sec ond$ at an angle of $35°$ to the horizontal. The ball leaves Brees’s hand at a height of $6feet$ .

Find parametric equations that model the position of the ball as a function of time.

Calculation:

The paramertric equations for the motion of the particle in projectile are given by the equations given as,

  $\begin{array}{l}x\left(t\right)=\left(s\cos \alpha \right)t\\ y\left(t\right)=-\frac{1}{2}g{t}^2+\left(s\sin \alpha \right)t+h\end{array}$

The initial speed given to the ball by the player is $s=80feet/\sec$ at angle of inclination with horizontal as $\theta =35°$ . The height of the ball when it leaves the player’s hand is $h=6feet$ .

So the equation is,

  $\begin{array}{l}x\left(t\right)=\left(80\cos 35°\right)t\\ y\left(t\right)=-\frac{1}{2}\times 32t^2+\left(80\sin 35°\right)t+6\end{array}$

Consider $y\left(t\right)=0$ ,

  $\begin{array}{l}-16t^2+\left(46\right)t+6=0\\ t=\frac{-46\pm \sqrt{{\left(46\right)}^2-4\left(-16\right)\left(6\right)}}{2\left(-16\right)}\\ t=\frac{-46\pm 50}{-32}\\ t=-\frac{1}{8},3\end{array}$

Consider positive time duration,

Hence, the position of the ball as a function of time $\boxed{3\sec onds}$

b.

To determine

How long is the ball in air?

Answer to Problem 83RE

  $\boxed{39.0625feet}$

Explanation of Solution

Given information:

Drew Brees throws a football with an initial speed of $80feet/\sec ond$ at an angle of $35°$ to the horizontal. The ball leaves Brees’s hand at a height of $6feet$ .

How long is the ball in air?

Calculation:

The maximum height of the ball can be determine by differentiating the height of the projectile and equation is to zero to obtain time instants, since the derivation of the height function expression the slope of the projectile path.

  $\begin{array}{l}y\text{'}\left(t\right)=0\\ -32t+46=0\\ t=\frac{23}{16}\end{array}$

Now derivative of this function,

  $y\text{'}\text{'}\left(t\right)=-32$

The double derivative of the height function is negative, the maximum height can be find by putting $t=\frac{23}{16}$ In the height function.

  $\begin{array}{l}y\left(t\right)=-16t^2+46t+6\\ =-16{\left(\frac{23}{16}\right)}^2+46\left(\frac{23}{16}\right)+6\\ =39.0625feet\end{array}$

Hence, the height of ball in air $\boxed{39.0625feet}$

c.

To determine

Determine the maximum height of the ball.

Answer to Problem 83RE

  $\boxed{39.0625feet}$

Explanation of Solution

Given information:

Drew Brees throws a football with an initial speed of $80feet/\sec ond$ at an angle of $35°$ to the horizontal. The ball leaves Brees’s hand at a height of $6feet$ .

When is the ball at its maximum height? Determine the maximum height of the ball.

Calculation:

The maximum height of the ball can be determine by differentiating the height of the projectile and equation is to zero to obtain time instants, since the derivation of the height function expression the slope of the projectile path.

  $\begin{array}{l}y\text{'}\left(t\right)=0\\ -32t+46=0\\ t=\frac{23}{16}\end{array}$

Now derivative of this function,

  $y\text{'}\text{'}\left(t\right)=-32$

The double derivative of the height function is negative, the maximum height can be find by putting $t=\frac{23}{16}$ In the height function.

  $\begin{array}{l}y\left(t\right)=-16t^2+46t+6\\ =-16{\left(\frac{23}{16}\right)}^2+46\left(\frac{23}{16}\right)+6\\ =39.0625feet\end{array}$

Hence, the maximum height of the ball is, $\boxed{39.0625feet}$

d.

To determine

Determine the horizontal distance the ball travels.

Answer to Problem 83RE

  $\boxed{196.5feet}$

Explanation of Solution

Given information:

Drew Brees throws a football with an initial speed of $80feet/\sec ond$ at an angle of $35°$ to the horizontal. The ball leaves Brees’s hand at a height of $6feet$ .

Determine the horizontal distance the ball travels.

Calculation:

The horizontal distance the ball travels,

  $\begin{array}{l}x\left(t\right)=\left(80\cos 35°\right)t\\ x\left(t\right)=65.5\times 3\left[\therefore t=3\right]\\ x\left(t\right)=196.5feet\end{array}$

Hence, the horizontal distance the ball travels $\boxed{196.5feet}$

e.

To determine

Using a graphing utility to graph the equation.

Answer to Problem 83RE

  

Explanation of Solution

Given information:

Drew Brees throws a football with an initial speed of $80feet/\sec ond$ at an angle of $35°$ to the horizontal. The ball leaves Brees’s hand at a height of $6feet$ .

Using a graphing utility to graph the equation.

  $\begin{array}{l}x\left(t\right)=\left(s\cos \alpha \right)t\\ y\left(t\right)=-\frac{1}{2}g{t}^2+\left(s\sin \alpha \right)t+h\end{array}$

Calculation:

Using a graphing utility to graph the equation is,

  $\begin{array}{l}x\left(t\right)=\left(s\cos \alpha \right)t\\ y\left(t\right)=-\frac{1}{2}g{t}^2+\left(s\sin \alpha \right)t+h\end{array}$

  

Hence the graph is drawn.