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
ft.

Answer 1

Set the heights equal because they meet at the same height:
-16t^2 + 56t = -16t^2 + 156t - 248.
The -16t^2 terms cancel, giving 56t = 156t - 248 ⇒ 100t = 248 ⇒ t = 2.48 s (after the first ball was hit).

Plug t into either height equation:
h = -16(2.48)^2 + 56(2.48) = 40.4736 ft ≈ 40.5 ft.

The balls meet at a height of 40.5 ft.