Example 5 from Algebra and Trigonometry, 10.7 Parametric Equations: Graphs

Finding the Parametric Equations to Describe the Motion of a Baseball

Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of $45°$ to the horizontal, making contact 3 feet above the ground.

ⓐ Find the parametric equations to model the path of the baseball.
ⓑ Where is the ball after 2 seconds?
ⓒ How long is the ball in the air?
ⓓ Is it a home run?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐ
  Use the formulas to set up the equations. The horizontal position is found using the parametric equation for $x .$ Thus,
  
  $\begin{array}{l} x = (v_{0} \cos θ) t \\ x = (140 \cos (45°)) t \end{array}$
  The vertical position is found using the parametric equation for $y .$ Thus,
  
  $\begin{array}{l} y = - 16 t^{2} + (v_{0} \sin θ) t + h \\ y = - 16 t^{2} + (140 \sin (45°)) t + 3 \end{array}$
2. ⓑ
  Substitute 2 into the equations to find the horizontal and vertical positions of the ball.
  $\begin{array}{l} x = (140 \cos (45°)) (2) \\ x = 198 feet \\ y = - 16 {(2)}^{2} + (140 \sin (45°)) (2) + 3 \\ y = 137 feet \end{array}$
  After 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground.
3. ⓒ
  To calculate how long the ball is in the air, we have to find out when it will hit ground, or when $y = 0.$ Thus,
  
  $\begin{array}{l} y = - 16 t^{2} + (140 \sin (45^{\circ})) t + 3 \\ y = 0 & Set y (t) = 0 and solve the quadratic . \\ t = 6.2173 \end{array}$
  When $t = 6.2173$ seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.)
4. ⓓ
  We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for $t,$ and input $t$ into $y .$
  $\begin{array}{l} x = (140 \cos (45°)) t \\ 400 = (140 \cos (45°)) t \\ t = 4.04 \end{array} \begin{array}{l} y = - 16 {(4.04)}^{2} + (140 \sin (45°)) (4.04) + 3 \\ y = 141.8 \end{array}$
  The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See Figure 7.
Figure 7

How did it go?

Solution

Solution