SOLUTION: A stone was thrown from the top of a cliff 60 meters above sea level. The height of the
stone above sea level t seconds after it was released, is given by H(t) = -5t2 + 20t + 60
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-> SOLUTION: A stone was thrown from the top of a cliff 60 meters above sea level. The height of the
stone above sea level t seconds after it was released, is given by H(t) = -5t2 + 20t + 60
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Question 1195857: A stone was thrown from the top of a cliff 60 meters above sea level. The height of the
stone above sea level t seconds after it was released, is given by H(t) = -5t2 + 20t + 60
meters.
a) Find the time it takes for the stone to reach its maximum height.
b) Find the maximum height above sea level reached by the stone.
c) How long did it take before the stone struck the water? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A stone was thrown from the top of a cliff 60 meters above sea level. The height of the stone above sea level t seconds after it was released, is given by H(t) = -5t^2 + 20t + 60 meters.
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a) Find the time it takes for the stone to reach its maximum height.
The max of the quadratic is at t = -b/2a
t = -20/-10 = 2 seconds
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b) Find the maximum height above sea level reached by the stone.
H(2) = -5*4 + 20*2 + 60 = 80 meters
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c) How long did it take before the stone struck the water?
--- Poorly worded question.
How many seconds passed from launch to impact?
H(t) = -5t^2 + 20t + 60 = 0
t^2 - 4t - 12 = 0
(t-6)*(t+2) = 0
t = 6 seconds to impact --- ignore the negative solution
