Question

A particle drops from a height of 10m on sand and stops after 1m. what is the retardation provided by sand

Answer 1

let the height be h .

(u is zero since its dropped.)

so the time taken to reach the ground is:

s=ut+1/2at^2

10=1/2*10*t*t

t=2^1/2 or 1.414

velocity at the end of journey (when it just strikes the ground).

v=u+at

v=10*1.414

v=14.14

now we know that it goes 1m into the sand with retardation ,so

s=1

a=a

u=14.14

v=0

by v^2-u^2=2as

o-14.41^2=2*a*1

a=100

hope the answer is correct.