Asked by Anonymous
An object is propelled vertically upward from the top of a 144-foot building. The quadratic function s(t)= -16t2+192+144 models the ball's height above the ground, in feet, s(t) seconds after it was thrown. How many seconds does it take until the object finally hits the ground? Round to the nearest tenth of a second if necessary.
Answer
0.7 seconds
Answer
0.7 seconds
Answers
Answered by
Reiny
when it hits the ground, s(t) = 0
-16t^2 + 192t + 144 = 0
t^2 - 12t - 9 = 0
I will complete the square rather than use the quadratic formula
t^2 - 12t + 36 = 9 + 36
(t-6)^2 = 45
t-6 = ±√45
t = 6 ± √45
= appr -.7 or 12.7
since t > 0, it will hit the ground 12.7 after it was tossed.
-16t^2 + 192t + 144 = 0
t^2 - 12t - 9 = 0
I will complete the square rather than use the quadratic formula
t^2 - 12t + 36 = 9 + 36
(t-6)^2 = 45
t-6 = ±√45
t = 6 ± √45
= appr -.7 or 12.7
since t > 0, it will hit the ground 12.7 after it was tossed.
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