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?
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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
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-4.9x^2 + 20x + 100 - 80 = 0
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-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
{{{x = (-20 +- sqrt( 20^2 - 4 * -4.9 * 20 ))/(2*-4.9) }}}
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{{{x = (-20 +- sqrt( 400 - (-392)))/(-9.8) }}}; minus a minus = +
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{{{x = (-20 - sqrt( 792))/(-9.8) }}}
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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:
{{{ graph( 300, 200, -2, 8, -10, 125, -4.9x^2 + 20x + 100) }}}
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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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Did this make sense to you?