souda
 Subject: mechanics Sun Sep 06, 2009 12:50 pm  
 a body is thrown vertically upward with a velocity u. it will fall back on a point displaced to the west.find it.let h be the hight.h=u^2/2g 

QUEST Active member
 Subject: Re: mechanics Sun Sep 06, 2009 10:13 pm  
 casea] when u throw a body vertically up it falls back to the same point under uniform gravity force assuming no other external force(s) acting on it.
caseb] assuming body to fall back at a displaced point x distance west of the initial point, it will be possible only when a body is thrown at an angle b and no other forces are acting on it except gravity. then ,
x=2ucos{b}t, where t is the time it takes to reach max. height h.
now, you value for h=u^2/2g suggest that the body has been thrown vertically up at an angle b=90 deg. proof: v^2=u^2  2gh so, 0 = u^2sin^2{b} 2g*u^2/2g = u^2[sin^2{b}  1] => sin^2{b} = 1 or, sin{b}= 1 or sin{b}= 1 i.e., b=90 deg or 270 deg which means vertical throw and hence proves that with the given data the body will fall back to its initial position in time 2t and x=0 b'cose cos 90 = 0 _________________ Character is like a tree and reputation like its shadow. The shadow is what we think of it; the tree is the real thing. Abraham Lincoln, Lincoln's Own Stories 16th president of US (1809  1865) 

souda
 Subject: Re: mechanics Sun Sep 06, 2009 11:17 pm  
 :)thanks for ur answer i think it is not correct.b'cos earth moving with a velocity omega(w)it must affect the motion.a freely falling body displaced at a distance x=1/3wg[2h/g]^3/2cos{b}.where b is the latitude.i have the answer for my question.bt i dont know how it is.the answer is x=4/3[8h^3/g]^1/2wcos{b}.you know how,then pls help me 

MADANKABUL
 Subject: Re: mechanics Wed Nov 03, 2010 12:26 pm  
 I NEED SOME TIPS IN EMI 
