Given:
Initial velocity of the body is u
At point A :
v1 =
At point B :
v2 =
At point C :
v3 =
As the body is moving upwards, acceleration a = - g (acceleration due to gravity)
Let the body be thrown from point O. Thus,
v1,v2 and
v3 are the final velocities at the end of distances OA , OB and OC respectively.
The corresponding figure is shown below :
Using the third equation of motion :
v2−u2 = 2aS
For distance OA:
v12−u2 = - 2g (OA)
⇒ OA =
=
for distance OB :
v22−u2 = - 2g (OB)
⇒ OB =
=
for distance OC :
v32−u2 = - 2g (OC)
⇒ OC =
=
Thus,
AB = OB - OA
⇒ AB =
− =
And
BC = OC - OB
⇒ BC =
− =
Therefore,
=
=
Hence, option 'D' is correct.