1
advanced-math

A ball is thrown upward with initial velocity 48 ft/s from a height of 6 ft. The height is h(t)=−16t2+48t+6h(t) = -16t^2 + 48t + 6. What is the maximum height?

A

42 ft

B

6 ft

C

1.5 ft

D

54 ft

Correct Answer: A

Choice A is the correct answer.

  1. Find Time to Max: t=−b/2a=−48/(2(−16))=1.5t = -b/2a = -48 / (2(-16)) = 1.5 seconds.
  2. Find Max Height: Substitute t=1.5t=1.5 into h(t)h(t). h(1.5)=−16(1.5)2+48(1.5)+6h(1.5) = -16(1.5)^2 + 48(1.5) + 6h(1.5)=−16(2.25)+72+6h(1.5) = -16(2.25) + 72 + 6h(1.5)=−36+72+6=42h(1.5) = -36 + 72 + 6 = 42 ft.

Choice B is incorrect because it is the initial height. Choice C is incorrect because it is the time. Choice D is incorrect.

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Set 17: Quadratic Equations and Functions 17