Velocity Of The Ice As It Hits The Ground?

AI Thread Summary
The discussion centers on calculating the velocity of a chunk of ice falling from the John Hancock Center, which is 343 meters tall. The initial velocity is zero, and the acceleration due to gravity is considered at 9.8 m/s². Using the formula Vf² = Vi² + 2as, the calculation leads to a final velocity of approximately 82 m/s upon impact. A correction is noted regarding the sign of gravity, emphasizing that it should be treated as positive in this context. The conversation highlights the application of basic physics principles to determine the impact velocity of falling objects.
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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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