SOLUTION: I really need help with this question. When a ball is thrown vertically upward, its height (h, in feet)above the ground after t seconds is described by the mathematical formula

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Question 60892: I really need help with this question.
When a ball is thrown vertically upward, its height (h, in feet)above the ground after t seconds is described by the mathematical formula
h=12t^2 + 96t +80. At what times is the ball 103 feet above the ground?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for expressing the height, h, at time, t, of an object thrown upwards with an initial velocity of Vo from an initial height of Ho, is given by:
f%28h%29+=+-16t%5E2+%2BVot+%2B+Ho (Notice that the coefficient of t^2 is negative to account for the doward (negative) effect of gravity.
In your problem, the initial upward velocity, Vo = 96 and the initial height, Ho = 80.
You want to find at what time, t, will f(h) = 103, so we'll set the equation = 103 and solve for t.
-16t%5E2%2B96t%2B80+=+103 Subtract 103 from both sides.
-16t%5E2%2B96t-23+=+0 Solve this quadratic equation using the quadratic formula: t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a where: a = -16, b = 96, and c = -23, and because you are solve a quadratic equation, you can expect to get two answers for t.
-16t%5E2%2B96t-23+=+0
t+=+%28-96%2B-sqrt%2896%5E2-4%28-16%29%28-23%29%29%29%2F2%28-16%29
t+=+%28-96%2B-sqrt%289216-1472%29%29%2F-32
t+=+%28-96%2B-sqrt%287744%29%29%2F-32
t+=+%28-96%2Bor-88%29%2F-32
t+=+3%2B%28-2.75%29 = 0.25
t+=+3-%28-2.75%29 = 5.75
The answers are:
t = 5.75 seconds. The ball is at 103 feet on the way down.
t = 0.25 seconds. The ball is at 103 feet on the way up.