UPSEE Solved Model Paper 7

Given:
Initial velocity of the body is u
At point A : v1 =

u

2

At point B : v2 =

u

3

At point C : v3 =

u

4

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 =

(u−2)2−u2

−2g

=

3u2

8g

for distance OB :
v22−u2 = - 2g (OB)
⇒ OB =

( u 3 )2−u2

−2g

=

8u2

18g

for distance OC :
v32−u2 = - 2g (OC)
⇒ OC =

( u 4 )2−u2

−2g

=

15u2

32g

Thus,
AB = OB - OA
⇒ AB =

8u2

18g

−

3u2

8g

=

5u2

72g

And
BC = OC - OB
⇒ BC =

15u2

32g

−

8u2

18g

=

7u2

288g

Therefore,

AB

BC

=

5u2 72g

7u2 288g

=

20

7

Hence, option 'D' is correct.