SOLUTION: An object is thrown upward from the top of a 160​-foot building with an initial velocity of 144 feet per second. The height h of the object after t seconds is given by the q
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Question 1066936: An object is thrown upward from the top of a 160-foot building with an initial velocity of 144 feet per second. The height h of the object after t seconds is given by the quadratic equation h= −16t^2+144t+160. When will the object hit the ground?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An object is thrown upward from the top of a 160-foot building with an initial velocity of 144 feet per second.
The height h of the object after t seconds is given by the quadratic equation h= −16t^2+144t+160.
When will the object hit the ground?
:
When the objects hits the ground, h=0 therefore we can write this:
-16t^2 + 144t + 160 = 0
Solve for t
We can simplify this, divide thru by -16, resulting in:
t^2 - 9t - 10t = 0
Factors easily to
(t-10)(t+1) = 0
The positive solution
t = 10 seconds is the time from thrown to hitting the ground
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