SOLUTION: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m?

Question 83818: A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A ball is thrown upward from the roof of a building 100 m tall with an initial velocity of 20 m/s. When will the ball reach a height of 80 m?
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The equation for this: h(x) height and x = time in seconds
h(x) = -4.9x^2 + 20x + 100
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To find time when ball is 80 m above the ground:
-4.9x^2 + 20x + 100 = 80
:
-4.9x^2 + 20x + 100 - 80 = 0
:
-4.9x^2 + 20x + 20 = 0
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Solve for x to find the time (in seconds) it will be 80 ft above the ground
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Use the quadratic formula; a=-4.9, b=20, c=20

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; minus a minus = +
:

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Do the math here you should get a positive solution of;
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x = 4.9 seconds for it to be at 80 m
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A graphical presentation of this event:

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Notice that when x = 5 sec, it will be at about 80 meters (y)
Also notice that it will be 20 meters below the height of the building (x=0)
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