A body is dropped from a height of 2m. It penetrates into the sand on the ground through a distance of 10cm before coming to rest. What is the retardation of the body in the sand? Please explain clearly.

Asked by Shourya Mukherjee | 22nd May, 2012, 07:05: PM

Expert Answer:

When the body drops from height 2m & will reach the ground, its  u = 0 
v = ?
 
v2 - u2 = 2gh
 
v2 = 2 * 9.8 * 2 = 4 * 9.8
 
 
Now, this v is the initial velocity for calculation retardation
And final velocity is zero
 
And distnace s = 10 cm
 
from the formula  v2 - u2 = 2as
 
a = v2-u2 /2s
 
a = -196 m/s2
 
 

Answered by  | 23rd May, 2012, 10:51: AM

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