Question

An object is thrown upward from the top of a 192 ft building with an initial velocity of 64 feet

per second. The height, h, of the object after, t seconds, is given by the equation h =
- 16t2 + 640 + 192 When will the object hit the ground?

Answer 1

t=6

Explanation:

ground height = 0

(are you sure your formula is correct? isn't it - 16t²?)

if h=16t² +64t+192 is true then

16t² +64t+192 = 0

t² + 4t + 12 = 0

t = (-4 ± √(4² - 4*12)) / 2*1 = (-4 ± √-32) / 2 = -2 ± 2√-2

There is no solution of t

if it is h= - 16t² +64t+192

0 =- 16t² + 64t + 192

t² - 4t - 12 = 0

(t + 2) (t -6) = 0

t should be positive

t = 6 sec