Velocity Of The Ice As It Hits The Ground?

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SUMMARY

The discussion centers on calculating the velocity of a chunk of ice falling from the John Hancock Center in Chicago, which stands at 343 meters. The initial velocity (Vi) is 0 m/s, and the acceleration due to gravity (A) is 9.8 m/s². Using the formula Vf² = Vi² + 2as, the final velocity (Vf) is calculated to be 82 m/s. A common mistake noted in the discussion is the incorrect use of negative gravity, which should be positive as the object accelerates towards the Earth.

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luvzdaladeez
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The John Hancock Center In Chicago Is The Tallest Building In The United States In Which There Are Residential Apartments. The Hancock Center Is 343m Tall. Suppose A Resident Accidentally Causes A Chunk Of Ice To Fall From The Roof. What Would Be The Velocity Of The Ice As It Hits The Ground? Neglect The Air Resistance
 
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Seem pretty stright forward, what is your question?
 
luvzdaladeez said:
The John Hancock Center In Chicago Is The Tallest Building In The United States In Which There Are Residential Apartments. The Hancock Center Is 343m Tall. Suppose A Resident Accidentally Causes A Chunk Of Ice To Fall From The Roof. What Would Be The Velocity Of The Ice As It Hits The Ground? Neglect The Air Resistance


1) What did you try?
2) See above
3) Think energy
 
Ok I Tried


D=343m
Vi=0
A= -9.8 M/s


Looking For Vf=


Vf^2=0^2+2 X -9.8 X 343

And Then I Solved It From There
 
Looks like.
 
s=343 u=0 a=9.8


v^2=u^2+2as

v^2=2as

v^2=2*343*9.8
v^2=6722.8
v=82 m/s

the mistake you've made is putting gravity as -9.8 when it is actually 9.8 as the object is accelerating towards the earth.
 

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