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A particle is thrown Ā vertically upwards.Its velocity at one fourth of maximum height is 20m/s.Then the max height is
let H is maximum height ... if initial velocity is u then applying , v2 = u2 + 2as at maximum height , v=0 , s=H , a=-g ,after putting these 0 = u2 -2gH or u2 = 2gH ...................1 again applying , v2 = u2 + 2as upto that instant when its velocity was 20m/s... s = H/4 , v = 20m/s 202 = u2 - 2g(H/4) .........2 from 1 putting u2 in 2 202 = 2gH - 2g(H/4) 400 = 2gH/3 H=80/3 = 26.7m approve if u like my ans
let H is maximum height ...
if initial velocity is u then
applying , v2 = u2 + 2as
at maximum height , v=0 , s=H , a=-g ,after putting these
0 = u2 -2gH or u2 = 2gH ...................1
again applying , v2 = u2 + 2as upto that instant when its velocity was 20m/s...
s = H/4 , v = 20m/s
202 = u2 - 2g(H/4) .........2
from 1 putting u2 in 2
202 = 2gH - 2g(H/4)
400 = 2gH/3
H=80/3 = 26.7m
approve if u like my ans
if the maximum height is H then at one-fourth of this height v^2=u^2-2gH/4, v=20m/s 400+2gH/4=u^2 .....1 now at the heighest point v=0 v^2=u^2-2gH 0=u^2-2gH u^2=2gH putting value of u^2 from equation 1 400+2gH/4=2gH H=80/3 m
if the maximum height is H then at one-fourth of this height
v^2=u^2-2gH/4, v=20m/s
400+2gH/4=u^2 .....1
now at the heighest point v=0
v^2=u^2-2gH
0=u^2-2gH
u^2=2gH
putting value of u^2 from equation 1
400+2gH/4=2gH
H=80/3 m
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