Chapter 6, Problem 79RE

Solution

a.

To write an equation that models the height of the arrow as a function of $t$ .

  $h\left(t\right)=-16t^2+245t+200$

Given information:

A person shoots the arrow with an initial velocity 245 ft/sec straight up. The arrow leaves from a point 200 ft above the ground level.

Calculation:

Since the initial velocity is 245 and the arrow leaves from a height of 200 ft, the equation modeling the height of the arrow at any time $t$ can be given as:

  $\begin{array}{l}h\left(t\right)=-\frac{1}{2}\cdot 32t^2+245t+200\\ h\left(t\right)=-16t^2+245t+200\end{array}$

Conclusion:

The required equation that models the height of the arrow is $h\left(t\right)=-16t^2+245t+200$ .

b.

To write a parametric equation to simulate the height of the flight of the arrow.

The parametric equation that simulates the height of the flight of the arrow is $x\left(t\right)=t$ , $y\left(t\right)=-16t^2+245t+200$ .

Given information:

A person shoots the arrow with an initial velocity 245 ft/sec straight up. The arrow leaves from a point 200 ft above the ground level.

Calculation:

The equation modeling the height of the arrow at any time $t$ can be given as $h\left(t\right)=-16t^2+245t+200$ .

The parametric equation that simulates the height of the flight can be given as:

  $x\left(t\right)=t$ , $y\left(t\right)=-16t^2+245t+200$

Conclusion:

The parametric equation that simulates the height of the flight of the arrow is $x\left(t\right)=t$ , $y\left(t\right)=-16t^2+245t+200$ .

c.

To graph the parametric equation that models the height of the arrow.

Given information:

A person shoots the arrow with an initial velocity 245 ft/sec straight up. The arrow leaves from a point 200 ft above the ground level.

Calculation:

The parametric equation that simulates the height of the flight can be given as:

  $x\left(t\right)=t$ , $y\left(t\right)=-16t^2+245t+200$

Use a graphing calculator to graph the equation as shown below.

  

d.

To find the heigh of the arrow after 4 sec.

The parametric equation that simulates the height of the flight of the arrow is $x\left(t\right)=t$ , $y\left(t\right)=-16t^2+245t+200$ .

Given information:

A person shoots the arrow with an initial velocity 245 ft/sec straight up. The arrow leaves from a point 200 ft above the ground level.

Calculation:

The equation modeling the height of the arrow at any time $t$ can be given as $h\left(t\right)=-16t^2+245t+200$ .

Substitute $t=4$ in $h\left(t\right)=-16t^2+245t+200$ to find the heigh of the arrow after 4 sec.

  $\begin{array}{c}h\left(4\right)=-16{\left(4\right)}^2+245\left(4\right)+200\\ =-16\cdot 16+980+200\\ =924\end{array}$

Conclusion:

The height of the arrow after 4 sec is 924 ft.

e.

To find the time when the arrow hits the ground.

  $16.0894$ sec.

Given information:

A person shoots the arrow with an initial velocity 245 ft/sec straight up. The arrow leaves from a point 200 ft above the ground level.

Calculation:

The equation modeling the height of the arrow at any time $t$ can be given as $h\left(t\right)=-16t^2+245t+200$ .

When arrow hits the ground, $h\left(t\right)=0$ .

So, solve the equation $-16t^2+245t+200=0$ .

  $\begin{array}{c}-16t^2+245t+200=0\\ t=\frac{-245\pm \sqrt{{245}^2-4\left(-16\right)\cdot 200}}{2\left(-16\right)}\\ t=\frac{-245\pm 5\sqrt{2913}}{2\left(-16\right)}\\ t=\frac{-245+5\sqrt{2913}}{32},t=\frac{245+5\sqrt{2913}}{32}\\ t=-0.77690,t=16.08940\end{array}$

Since the cannot be negative, the value $t=-0.77690$ is discarded.

Conclusion:

The arrow takes $t=16.0894$ sec to hit the ground.