Velocity Of The Ice As It Hits The Ground?

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Homework Help Overview

The discussion revolves around a physics problem involving the calculation of the velocity of a chunk of ice falling from the John Hancock Center, which is 343 meters tall. The problem specifies neglecting air resistance.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial conditions of the problem, including height and initial velocity. Some suggest using energy principles, while others attempt to apply kinematic equations to find the final velocity.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the assumptions made about gravitational acceleration. There is a mix of approaches being explored, with some participants noting potential errors in the application of gravity's value.

Contextual Notes

There is an emphasis on the conditions of the problem, particularly the neglect of air resistance and the initial velocity being zero. The discussion reflects varying interpretations of the gravitational acceleration's sign in the equations used.

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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