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PostPosted: Mon, 21 May 2012 04:35:43 UTC 
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A hawk flying at 11 m/s at an altitude of 132 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation:

y = 132 − (x^2/33)

until it hits the ground, where y is its height above the ground and x is its horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.

http://i1084.photobucket.com/albums/j40 ... IMG009.png


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PostPosted: Mon, 21 May 2012 05:28:05 UTC 
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Your work looks great. But your numerical answer is off. I get

$-\frac{33}2\left(-2\sqrt{17}+\frac12\ln\left|\sqrt{17}-4\right|\right)

$=\frac{33}4\left[4\sqrt{17}-\ln\left(\sqrt{17}-4\right)\right]

$\approx153.344\mathrm{\ m}

I think you got the parentheses wrong around the second square root when you entered the expression into your calculator. For if I replace \sqrt{17}-4 with \sqrt{17-4} in the above expression, I too get the erroneous answer of 125.48.


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PostPosted: Mon, 21 May 2012 16:45:17 UTC 
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Reckoner wrote:
Your work looks great. But your numerical answer is off. I get

$-\frac{33}2\left(-2\sqrt{17}+\frac12\ln\left|\sqrt{17}-4\right|\right)

$=\frac{33}4\left[4\sqrt{17}-\ln\left(\sqrt{17}-4\right)\right]

$\approx153.344\mathrm{\ m}

I think you got the parentheses wrong around the second square root when you entered the expression into your calculator. For if I replace \sqrt{17}-4 with \sqrt{17-4} in the above expression, I too get the erroneous answer of 125.48.


That was it! Thank you. :!:


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