A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball's height above
ground can be modeled by the equation H (t) = -16t^2+ 80t + 40.

1) The ball is thrown from a height of 40 feet, how long does it take the ball to fall back to
that height?
A 2.5
C 140
B 5
D 5.46

2) How much time passes when the ball reaches its maximum height?
A. 25
B.5
C.140
D.5.45

Question

contestada

Respuesta :

raphealnwobi raphealnwobi · Answer 1 · 2021-11-10T10:29:47+01:00

it takes 5 seconds to reach a height of 40 ft. and It takes 2.5 seconds to reach maximum height.

Let h(t) represent the height of the ball at time t.

Given that the height is given by:

h(t) = -16t² + 80t + 40

1) For the ball to reach a height 0f 40 ft, h(t) = 40, hence:

40 = -16t² + 80t + 40

16t² - 80t = 0

t(t - 5) = 0

t = 0 or t = 5

Hence it takes 5 seconds to reach a height of 40 ft.

2) The maximum height is at h'(t) = 0,

h'(t) = -32t + 80

-32t + 80 = 0

32t = 80

t = 2.5

It takes 2.5 seconds to reach maximum height.

Find out more at: https://brainly.com/question/11535666