mechanics

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

Active member

case-a] 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.

case-b] 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

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

Subject: Re: mechanics Wed Nov 03, 2010 12:26 pm

I NEED SOME TIPS IN EMI