The height, in feet, after x seconds of an object launched straight up can be found by the function h (x) = - 16x2+v0x+h0, where v0 is the initial velocity of the object and h0 is the initial height.

A ball is kicked straight up into the air from a height of 12 ft with an initial velocity of 44 ft/s. After how many seconds does the ball hit the ground?

Enter your answer in the box.

___ s

Question

Leroy Arroyo

Mathematics

22 July, 12:43

The height, in feet, after x seconds of an object launched straight up can be found by the function h (x) = - 16x2+v0x+h0, where v0 is the initial velocity of the object and h0 is the initial height.

A ball is kicked straight up into the air from a height of 12 ft with an initial velocity of 44 ft/s. After how many seconds does the ball hit the ground?

Enter your answer in the box.

___ s

+3

Answers (1)

Know the Answer?

Kelvin Newton · Answer 1

The function is:

h (x) = - 16 x² + vo x + h o

v o = 44 ft / s (initial velocity)

h o = 12 ft (initial height)

- 16 x² + 44 x + 12 = 0 / : 4 (we will divide both sides by 4)

- 4 x² + 11 x + 3 = 0

- 4 x² + 12 x - x + 3 = 0

- 4 x (x - 3) - (x - 3) = 0

(x - 3) ( - 4 x - 1) = 0

x - 3 = 0, x = 3 s

or:

- 4 x - 1 = 0, - 4 x = 1, x = - 1/4 (false).

Answer:

3 seconds.