SOLUTION: Quinn punted the football from a height of 3 feet above the turf field.The ball was kicked with an initial upward velocity of 68 feet per second Will the ball reach a height of 70

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Question 954172: Quinn punted the football from a height of 3 feet above the turf field.The ball was kicked with an initial upward velocity of 68 feet per second
Will the ball reach a height of 70 feet?
How many seconds until the ball hits the ground?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Quinn punted the football from a height of 3 feet above the turf field.The ball was kicked with an initial upward velocity of 68 feet per second
Will the ball reach a height of 70 feet?
h(t) = -16t^2 + v*t + h0 is commonly used for ballistics problems.
h(t) = -16t^2 + 68t + 3
-16t^2 + 68t + 3 = 70 Solve for t
-16t^2 + 68t - 67 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B68x%2B-67+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2868%29%5E2-4%2A-16%2A-67=336.

Discriminant d=336 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-68%2B-sqrt%28+336+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2868%29%2Bsqrt%28+336+%29%29%2F2%5C-16+=+1.55217803813052
x%5B2%5D+=+%28-%2868%29-sqrt%28+336+%29%29%2F2%5C-16+=+2.69782196186948

Quadratic expression -16x%5E2%2B68x%2B-67 can be factored:
-16x%5E2%2B68x%2B-67+=+%28x-1.55217803813052%29%2A%28x-2.69782196186948%29
Again, the answer is: 1.55217803813052, 2.69782196186948. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B68%2Ax%2B-67+%29

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t =~ 1.552 seconds ascending
t =~ 2.698 seconds descending
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How many seconds until the ball hits the ground?
Use the same equation, find h(t) = 0
h(t) = -16t^2 + 68t + 3 = 0