SOLUTION: how many seconds will the ball be 75 ft haigh after hit 2 feet off the ground and initial upward velocity of 90 ft / sec. 75h = 2 ho + 90 t - 16 t squared

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: how many seconds will the ball be 75 ft haigh after hit 2 feet off the ground and initial upward velocity of 90 ft / sec. 75h = 2 ho + 90 t - 16 t squared      Log On


   

Question 388044: how many seconds will the ball be 75 ft haigh after hit 2 feet off the ground and initial upward velocity of 90 ft / sec.
75h = 2 ho + 90 t - 16 t squared
Found 2 solutions by solver91311, Alan3354:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You have exactly the right idea:





Solve the quadratic using the quadratic formula. You will find two zeros as is typical with quadratic. The first root is the time when the ball reaches 75 feet on the way up. The second root is the time when the ball reaches 75 feet high on the way down.


John

My calculator said it, I believe it, that settles it
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Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
how many seconds will the ball be 75 ft haigh after hit 2 feet off the ground and initial upward velocity of 90 ft / sec.
75h = 2 ho + 90 t - 16 t squared
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h(t) = -16t^2 + 90t + 2
At h = 75:
-16t^2 + 90t + 2 = 75
-16t^2 + 90t - 73 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B90x%2B-73+=+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=%2890%29%5E2-4%2A-16%2A-73=3428.

Discriminant d=3428 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-90%2B-sqrt%28+3428+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2890%29%2Bsqrt%28+3428+%29%29%2F2%5C-16+=+0.982839853961944
x%5B2%5D+=+%28-%2890%29-sqrt%28+3428+%29%29%2F2%5C-16+=+4.64216014603806

Quadratic expression -16x%5E2%2B90x%2B-73 can be factored:
-16x%5E2%2B90x%2B-73+=+%28x-0.982839853961944%29%2A%28x-4.64216014603806%29
Again, the answer is: 0.982839853961944, 4.64216014603806. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B90%2Ax%2B-73+%29

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The ball passes 75 feet twice, ascending at t = 0.982839853961944, and descending at t = 4.64216014603806 seconds.
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The ball is exactly 75 feet high for zero time, or dt seconds.
It is at or above 75 feet for the difference between the 2 times above,
=~ 3.659 seconds