when a body is projected vertically up from the ground its velocity is reduced to 1/4th of its velocity at ground .then the maximum height reached by the body is
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let initial velocity be u.
then, final velocity v is u/4.
Using the equation ,
v^2= u^2+2as
u^2/4= u^2- 2×10× h
2×10× h= u^2- u^2/4
2×10×h = 3u^2/4
h= 3× u^2/ 80
then, final velocity v is u/4.
Using the equation ,
v^2= u^2+2as
u^2/4= u^2- 2×10× h
2×10× h= u^2- u^2/4
2×10×h = 3u^2/4
h= 3× u^2/ 80
GayatriNaik:
but the options are:a#1/4 ,b# 4/3, c#2/3,d#3/5
Could you please check the question once again if the initial velocity is given..
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