A ball thrown upward and outward from a height of 6 feet. The height of the ball, f (x), in feet can be modelded by f (x) = - 0.8x^2 + 3.2x + 6, where x is the ball's horizontal distance, from where it was thrown.

a. what is the maximum height of the ball?

b. How far from where it was thrown does this height occur?

c. How far does the ball travel horizontally before hitting the ground? Round to the nearest tenth.

Question

Woodard

Mathematics

15 September, 03:37

A ball thrown upward and outward from a height of 6 feet. The height of the ball, f (x), in feet can be modelded by f (x) = - 0.8x^2 + 3.2x + 6, where x is the ball's horizontal distance, from where it was thrown.

a. what is the maximum height of the ball?

b. How far from where it was thrown does this height occur?

c. How far does the ball travel horizontally before hitting the ground? Round to the nearest tenth.

+4

Answers (1)

Know the Answer?

Renee Bowers · Answer 1

A. Convert to vertex form:-

= - 0.8 (x^2 - 4x) + 6

= - 0.8 ((x - 2) ^2 - 4)) + 6

= - 0.8 (x - 2) ^2 + 9.2

Maximum height of the ball is 9.2 feet Answer

b. horizontal distance at this height = 2 feet. Answer

c. this distance will be the positive root of f (x) = 0

-0.8 (x - 2) ^2 = - 9.2

(x - 2) ^2 = 11.5

xi-2 = sqrt 11.5 = 3.39

required distance = 5.39 = 5.5 feet answer