Chapter 6, Problem 86RE

Solution

To find:Will Spencer be able to do goal if the kick sails “true”?

Yes

Given information:

Spencer kicks the ball 40 yd away from the goal post, whose crossbar is 10 ft high.

Formula used:

Conversion formula from yard to feet, $1\text{yd}=3\text{ft}$ .

Calculation:

It is known that the path of an object in projectile motion is modelled by the parametric equations $x=\left(v_0\cos \theta \right)t$ and $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ , where $y_0$ is the point above the ground from where the object is thrown, $v_0$ is the initial speed in $\text{ft}/\sec$ at an angle $\theta$ with horizontal.

In the given situation, $v_0=70\text{ft}/\sec$ , $\theta =45°$ , and $y_0=0\text{ft}$ .

Substitute these values into the parametric equations $x=\left(v_0\cos \theta \right)t$ and $y=-16t^2+\left(v_0\sin \theta \right)t+y_0$ .

  $x=\left(70\cos 45°\right)t$ and $y=-16t^2+\left(70\sin 45°\right)t$

Evaluate the value of $t$ when $x=40\text{yd}$ or $x=120\text{ft}$ .

  $\begin{array}{c}120=\left(70\cos 45°\right)t\\ t=\frac{120}{70\cos 45°}\\ \approx 2.424\sec \end{array}$

Substitute $t\approx 2.424\sec$ into $y=-16t^2+\left(70\sin 45°\right)t$ and evaluate the value of $y$ .

  $\begin{array}{l}y=-16t^2+\left(70\sin 45°\right)\left(2.424\right)\\ y\approx 25.96\text{ft}\end{array}$

It is clear that $25.96$ is greater than 10 and 10 ft is the given hight of crossbar. So, Spencer makes the field goal.

Therefore, the answer is “Yes”.