Question 111036
The Height in feet of a baseball above ground is given by h(t)=-16t^2+32t+3 where t represents the time in seconds after the baseball is thrown.
If you throw the ball into the air, how many seconds have passed until the ball hits the ground?? 
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Look at the three elements of the equation:
-16t^2 is the force of gravity, negative because it is pulling it down
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+32t is the speed of the ball upward when it is thrown
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+3 is the point above the ground that the ball thrown
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t is in seconds
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h = height in feet
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When the ball hits the ground the height is 0,  so we have:
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-16t^2 + 32t + 3 = 0
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Use the quadratic formula to find t; a = -16; b = 32; c = 3
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
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{{{t = (-32 +- sqrt( 32^2 - 4* -16 * 3 ))/(2*-16) }}}
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{{{t = (-32 +- sqrt(1024 - (-192) ))/(-32) }}}
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{{{t = (-32 +- sqrt(1024 + 192 ))/(-32) }}}
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{{{t = (-32 +- sqrt(1216 ))/(-32) }}}
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{{{t = (-32 - 34.87)/(-32) }}}; we only want this solution
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{{{t = (-66.87)/(-32) }}}
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t = +2.089 sec, when it hits the ground
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A graph would look like this.   Vertical = height; horizontal = time in sec         
{{{ graph( 300, 200, -2, 4, -10, 25, -16x^2+32x+3) }}}
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Note that it crosses the y axis at 3' when t = 0 and max height is about 19 ft at 1 sec
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