An object is thrown from the top of an 80 foot building with an initial velocity of 64 feet per second. The height (h) of the object after (t) seconds is given by the quadratic equation h=-16t+64t+80 .When will the object hit the ground?

Answer 1

The 16t should be 16t^2. Check your Eq for errors.

d = 64t +16t^2 = 80
16t^2 + 64t - 80 = 0
(T-1)(T+5) = 0.
T = 1, and -s.

T = 1 s.

Answer 2

Correction:
16t^2 + 64t - 80 = 0
Divide both side by 16:
t^2 + 4t - 5 = 0.
(T-1)(t+5) = 0
T = 1, and -5.
Use positive value of T.

Answer 3

6.4