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
Messages
3
Reaction score
0
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
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top