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Vertical motion near the Earth's surface

 
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runner



Joined: 19 Apr 2011
Posts: 1
Location: australia

PostPosted: Wed Apr 20, 2011 5:28 am    Post subject: Vertical motion near the Earth's surface Reply with quote

Please help me - I'm sorry to say that I find the following question difficult to even decide on what formula to use or where to start.....

A surveryor's tape is hung vertically from the top of a building and a small ballbearing is dropped from the zero mark. A photograph of the falling ballbearing shows that it falls from the 4.90m mark to the 4.91m mark while the shutter of the camera is open. What is the length of the exposure?
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hepcj
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Joined: 23 Jun 2007
Posts: 125

PostPosted: Wed Apr 20, 2011 9:44 am    Post subject: Reply with quote

You are given the following information:
the start point x0= 0
the initial velocity u=0
a position where the camera shutter is opened x1 = 4.9m
a second position where the camera shutter closes. x2 = 4.91m

The motion is accelerated by gravity so we need to use a formula that relates distance, acceleration and time.

s = u t + 0.5 g t^2

since u= 0, s = 0.5 g t^2

so t = sqrt[(2s)/g]

s1 is the distance i.e. s = x1 - x0 = 4.9 m
s2 is the distance travelled when the shutter closes. s =x2-x0 = 4.91 m

calculate t for s1 and s2 and take the difference t2- t1.
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