SOLUTION: Suppose Charle O'Brian hits a baseball straight upward at 150 ft/sec from a height of 5 ft. Use the formula to determine how long it takes the ball to return to the earth.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Suppose Charle O'Brian hits a baseball straight upward at 150 ft/sec from a height of 5 ft. Use the formula to determine how long it takes the ball to return to the earth.       Log On


   

Question 124469This question is from textbook Elementary and Intermediate Algebra
: Suppose Charle O'Brian hits a baseball straight upward at 150 ft/sec from a height of 5 ft.
Use the formula to determine how long it takes the ball to return to the earth.
This question is from textbook Elementary and Intermediate Algebra

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The equation you are looking for is:
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h%28t%29+=+-16t%5E2+%2B+V%5Bo%5Dt+%2B+h%5Bo%5D
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V%5Bo%5D is the initial velocity of + 150 ft/s and h%5Bo%5D is the initial height of 5 ft.
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Substituting these values into the equation results in:
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h%28t%29+=+-16t%5E2+%2B+150t+%2B+5
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in which h%28t%29 represents the height of the ball above the ground at time t.
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You are interested in finding the time it takes for the ball to rise up as high as it will go, and
then finally fall back to the ground where its height will be zero feet. So substitute
zero for h%28t%29 and the equation becomes:
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0+=+-16t%5E2+%2B+150t+%2B+5
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which (by switching sides) in the more standard form of a quadratic equation is:
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-16t%5E2+%2B+150t+%2B+5+=+0
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This in standard form of at%5E2+%2B+bt+%2B+c in which a = -16, b = +150, and c = 5. The quadratic
formula tells you that for such standard form quadratic equations the answers for t will be
in the form:
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t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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Substitute the above values for a, b, and c and the equation becomes:
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t+=+%28-150+%2B-+sqrt%28+150%5E2-4%2A%28-16%29%2A5+%29%29%2F%282%2A%28-16%29%29+
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Determine the term under the radical. 150%5E2+=+22500 and %28-4%29%2A%28-16%29%285%29=320.
So the term under the radical becomes 22500+%2B+320+=+22820. Substitute this and the equation is:
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t+=+%28-150+%2B-+sqrt%28+22820%29%29%2F%282%2A%28-16%29%29+
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And sqrt%2822820%29+=+151.0629008
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Substitute this square root into the equation and you get:
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t+=+%28-150+%2B-+151.0629008%29%2F%282%2A%28-16%29%29
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The denominator multiplies out to -32 which makes the equation for t become:
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t+=+%28-150+%2B-+151.0629008%29%2F%28-32%29%29
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t must be a positive value (negative t doesn't make sense). Since the denominator is
negative, the numerator must also be negative to give a positive answer for t. The only way
to make the numerator negative is if both terms in the numerator are negative. Therefore, you
can forget about the + sign in the numerator and you only need to solve the equation:
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t+=+%28-150+-+151.0629008%29%2F%28-32%29%29
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Combining the two terms in the numerator results in:
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t+=+%28-301.0629008%29%2F%28-32%29
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and dividing the numerator by the denominator results in an answer of:
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t+=+9.40821565
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This rounds of to an answer of t+=+9.41 seconds.
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Hope this helps you to understand the problem.
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