SOLUTION: 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 a

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??

I really don't understand how to put this problem in the equations it's supposed to go into and I don't understand how to solve them...help!
Found 2 solutions by scott8148, ankor@dixie-net.com:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
when the ball hits the ground, the height (h(t)) is zero ___ so 0=-16t^2+32t+3

solve for t using the quadratic formula (the answer is between 2 and 3 seconds)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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??
:
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

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; we only want this solution
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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

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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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