Asked by Medusa.
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?
Answers
Answered by
Henry
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.
d = 64t +16t^2 = 80
16t^2 + 64t - 80 = 0
(T-1)(T+5) = 0.
T = 1, and -s.
T = 1 s.
Answered by
Henry
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.
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.
Answered by
Anonymous
6.4
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