Question

A body thrown upward from ground covers equal distance in 4th and 7th second . With what initial velocity body was projected ?

Answer 1

h=vi*t-1/2 g t^2
distance fourth second=vi*5-4.8*25-vi*4+4.8*16=vi-4.8(9)
distance seventh second= similar math=vi-4.8(64-49)=vi-4.8(15)
now, distances are "equal" if one ignores direction, but considering direction, then
vi-4.8*9=-(vi-4.8*15)
2vi=4.8(6)
vi= 2.4*6=14.4

check: distance in fourth second
h(5)-h(4)=14.4*(5-4)+4.8(16-25)
= 14.4+4.8*9=57.6
distance in seventh second
h(7)-h(8)=14.4(-1)+4.8(64-49)
= -14.4+4.8(15)=57.6

Answer 2

Distance in 4th second= Distance in 7th second.
Distance in 4th second= u+{a/2(2n-1)}...where n=4..
Solving this you will get
S=u-35...where a=-g=-10.(because body is moving upwards)
Same for n=7
You will get
S=u-65
Now distance will be same only when one at one time it is moving upwards and other time it is moving downwards..
So S4=-S7
U-35=-(U-65)
After solving you will get U=50m/s