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Pupil
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posted by  killian1981 on 11/1/2009 5:24:05 PM  |  status: Closed  |  Earned Karma: 50

only need help on 2nd part of question...life saver for 1st to answer..

Course Textbook Chapter Problem Needs by
Calculus Based Physics N/A N/A N/A 11/2/2009 at 11:00:00 AM
Question Details:
A bullet (m = 0.0230 kg) is fired with a speed of 92.00 m/s and hits a block (M = 2.50 kg) supported by two light strings as shown, stopping quickly. Find the height to which the block rises.
3.59×10-2 m
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Find the angle (in degrees) through which the block rises, if the strings are 0.220 m in length.
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Oracle
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posted by Altair on 11/1/2009 5:36:03 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
you alreasy got the height h = 3.59×10-2 m, let L be the length of the string
L = 0.220 m
the angle is: sinθ = (L-h)/L
θ = sin-1 ((L-h)/L) = sin-1 [(0.220m - 3.59*10-2)/0.220m] = 56.810
the angle is 56.810

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Oracle
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posted by redcloud173 on 11/1/2009 5:36:40 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
I can't see a picture, but I'm going to guess it looks like this:



if it doesn't look like that, send me a message and i'll fix it.

we know that L=.22 and h=.0359

by using trigonometry we can find the angle.  we know that the sum of the vertical component of the string length and the height h is equal to the total string length.  since θ is an angle from the vertical in my drawing, the vertical component of the string is given by Lcosθ:
feel free to send me a message if something isn't clear. and please rate me.
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