Question

A freely falling body crosses the points A, B, C with velocities v,2v,3v. Then AB:BC=​

Answer 1

Answer:

Let the acceleration be a.

For P to Q :

Initial speed  u=V

Final speed  v=2V

Using  v2−u2=2aS

∴  2aSPQ=(2V)2−V2

We get   SPQ=2a3V2

For Q to R :

Initial speed  u=2V

Final speed  v=3V

Using  v2−u2=2aS

∴  2aSQR=(3V)2−(2V)2

We get   SQR=2a5V2

So.   SQR:SPQ=3:5