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rphy (104)

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a ball is projected vertically upwards with a velocity (gR)1/2 , R is radius of earth. to what height will ball rise?

a)R
b)R/2
c)R/4
d)3R/4

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caprio.nups (171)

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is it R/2 ????

nupur..
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akku (1142)

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Hi rpy 
I think the ans is a)R
SOLUTION
by ENERGY CONSERVATION
TEi=TEf ---> PEi+KEi=PEf
-GmMe/R+1/2mv^2=-GmMe/(R+h) where Me=mass of earth and h=ht reached
g=Gme/R^2
-gR+1/2gR=-gR^2/(R+h) (since at heighest pt vel becomes zero)
=>h=R answer
 
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priyesh (1607)

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what's the answer????????
 

"Imagination is more important than knowledge."
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for_succes (294)

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 akku is rite:
n priyesh when u conserve energy u shd take into account the gravitational energy between the earth and the mass m coz we consider THE EARTH'S MASS TO BE CONCENTRATED AT ITS CENTRE OF MASS and now the mass m is situated at a distance of R frm the earth so akku's answer is justified

SOME SAY SKY IS THE LIMIT BUT TO ME
THERE'S SOMETHING BEYOND THE SKY
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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priyesh (1607)

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Thanks for_success for correcting my mistake.I've edited my post

"Imagination is more important than knowledge."
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ramyani (2899)

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Final vel = 0 (v)
initial vel = u = (gR)1/2
height= h

v^2 = u^2 --2gh  (u and g are opposite)
or, h =
u^2 / 2g = gR / 2g = R /2

option b is correct


it is not important where u stand, but in which direction u are moving
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rphy (104)

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yes thats the correct ans akku..

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karthik2007 (3733)

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Yeah, the answer is R. BTW, ramyani, the method that you've adopted to calculate the answer is totally wrong. You cannot use the equations of motion here as the acceleration due to gravity does not prevail during the entire motion of the body.

Will nip in at times to solve problems :)
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amitsingh (273)

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Answeris R. Methods are right.

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nibin89 (20)

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simple question .............equate the energy,,,,,,,,,,,,ans is (a)R
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ramyani (2899)

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thanks to all.i now got it right


it is not important where u stand, but in which direction u are moving
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