Q2.  A stone weighing 3kg from the top of a tower 100 meter high and burries itself 2m deep in the sand. the time of penetration is
 

Asked by vikasg13.hardware | 23rd Jun, 2017, 12:55: PM

Expert Answer:

Let us approach this interesting question with a step-by-step method:
 
1) The weight of the stone is a redundant value, not used in calculation.
2) At the time of falling, its initial velocity is u=0 m/s, the displacement s=100 and acceleration due to gravity g=-9.8 m/s2.
3) Applying above values in the formula, v2=u2+2as, we get the velocity when the stone just reaches the bottom of the tower and is about to penetrate the sand.
4) For penetration of stone inside the sand, the above calculated 'v' will now be initial velocity i.e 'u' and at the end of penetration, the final velocity v=0.
5) Applying above values in formula, v2=u2+2as, we get the acceleration 'a' of stone penetration in sand.
6) Using the initial velocity of sand penetration 'u', final velocity at the end of penetration v=0 and above calculated 'a' in sand in the formula v=u+at, we get the time taken to penetrate the sand by 2 m.
 

Answered by Abhijeet Mishra | 23rd Jun, 2017, 03:28: PM