SOLUTION: A ball is thrown upward from the roof of a building 100m tall with an initial velocity of 10m/s. When will the ball reach a hieght of 80m?
Same thing I can not understand how to
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Question 65746: A ball is thrown upward from the roof of a building 100m tall with an initial velocity of 10m/s. When will the ball reach a hieght of 80m?
Same thing I can not understand how to solve this type of problem, thank you so much for the help:-) Answer by Nate(3500) (Show Source):
You can put this solution on YOUR website! f(t) = at^2 + bt + c
a = -16
b = speed
c = height
t = time
f(t) = height after a unit of time (t)
80 = -16t^2 + 10t + 100
-20 = -16t^2 + 10t
20/16 = t^2 - 10t/16
20/16 = t^2 - 5t/8
20/16 + 25/16^2 = (t - 5/16)^2
320/16^2 + 25/16^2 = 345/16^2 = (t - 5/16)^2
+- sqrt(345)/16 = t - 5/16
5/16 +- sqrt(345)/16 = t
After about 1.4734 seconds, the ball will reach 80m.
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