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Old Jun 15th 2017, 01:32 AM   #1
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Question No.3

Hello again my friends

Could any one help me to find the suitable procedure to solve the attached problem ? i will thankful for him.

Best regards
Razi
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Old Jun 15th 2017, 06:35 AM   #2
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Use energy principles, as the problem suggested. The work done by the force of the wind equals the maximum potential energy gain of the ball. The work done by the wind is the integral oftegh force times the distance the ball travels in the direction of that force:

$\displaystyle W = \int _0 ^ \Theta \vec F \cdot d \vec S$

where S is the path length of the ball. Using $\displaystyle S = L \theta$, and then $\displaystyle \vec F \cdot d \vec S = |F| \cos \theta d \theta$, the work equation becomes:

$\displaystyle W = \int _0 ^ \Theta |F| \cos \theta d \theta $

where $\displaystyle \Theta$ is the maximum angle it can attain. Set this equal to the gain in PE:

$\displaystyle \Delta PE = mgH $

The rest is simply converting $\displaystyle \sin \Theta$ into units of L ang H, and manipulating the equations. Hope this helps.
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Old Jun 15th 2017, 09:21 AM   #3
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Dear ChipB

Thanks for replay , i have did your suggestion and i got this result :

H = F * L * sin(theta) / (m*g)

Now , how could eliminate sin(theta) ? and convert it to function of L ?

Thanks in advance and best regards
Razi
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Old Jun 15th 2017, 12:39 PM   #4
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$H = L-L\cos{\theta} \implies \cos{\theta} = \dfrac{L-H}{L}$

recall $\sin{\theta} = \sqrt{1-\cos^2{\theta}}$
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Old Jun 15th 2017, 01:11 PM   #5
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I did it , thank you sooooooooooooooooooooooooo much

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