SOLUTION: HELP ASAP!!!! An object is propelled vertically upward from the top of a 112-foot building. The quadratic function s(t) = -16t^2 + 176t + 112 models the ball’s height above t
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Question 80324: HELP ASAP!!!! An object is propelled vertically upward from the top of a 112-foot building. The quadratic function s(t) = -16t^2 + 176t + 112 models the ball’s height above the ground, s(t) in feet, 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.
Found 2 solutions by ankor@dixie-net.com, scott8148:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An object is propelled vertically upward from the top of a 112-foot building. The quadratic function s(t) = -16t^2 + 176t + 112 models the ball’s height above the ground, s(t) in feet, 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.
:
When the object hit's the ground the height, s(t) = 0 so we have:
:
-16t^2 + 176t + 112 = 0
:
Fortunately we can simplify things by dividing equation by -16
t^2 - 11t - 7 = 0
:
Use the quadratic formula: a=1; b=-11; c=-7 (we only care about the positive solution)
:
:
:
t = 11.6 sec
:
:
Check solution in original equation:
-16(11.6^2) + 176(11.6) + 112
-2153 + 2041 + 112 = 0
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the object hits the ground when the height s equals zero ... so set s=0 and solve for t
.. dividing by -16 gives .. no integer factors
using quadratic equation so ... t=-.6 and t=11.6 , negative value is not realistic
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