Question

During batting practice, two pop flies are hit from the same location, 2 s apart. The paths are modeled by the equations h = -16t2 + 56t and h = -16t2 + 156t - 248, where t is the time that has passed since the first ball was hit.

Explain how to find the height at which the balls meet. Then find the height to the nearest tenth.
The balls meet when their heights are equal: set
-16t^2 + 56t = -16t^2 + 156t - 248.
The -16t^2 terms cancel, leaving 56t = 156t - 248 ⇒ 100t = 248 ⇒ t = 2.48 s (after the first ball was hit).

Plug t into either height formula:
h = -16(2.48)^2 + 56(2.48) = -16(6.1504) + 138.88 = 40.4736 ft.

Rounded to the nearest tenth, they meet at 40.5 ft above the ground.
The balls meet at a height of
40.5
ft.
Which of these steps did you include in your solution? Check all of the boxes that apply.

Use substitution to write the equation
–16t2 + 56t = –16t2 + 156t – 248.

Simplify and solve for t, which is 2.48 s.

Substitute 2.48 for t into either equation to get h = 40.5 ft.

Answer 1

All three steps were included.

- Use substitution to write the equation –16t^2 + 56t = –16t^2 + 156t – 248. ✓
- Simplify and solve for t, which is 2.48 s. ✓
- Substitute 2.48 for t into either equation to get h = 40.5 ft. ✓