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A particle is thrown  vertically upwards.Its velocity at one fourth of maximum height is 20m/s.Then the max height is

8 years ago

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

8 years ago

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

8 years ago
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