Question #cdc14

1 Answer
Mar 4, 2016

Answer:

(B) #=(2+sqrt2)s#

Explanation:

Let #s# be the total distance covered in time #t#.
We know from general expression that #s=ut+1/2g.t^2#
Since the object falls from rest, #u=0#
Equation becomes #s=1/2g.t^2# ......(1)

It is given that object covers half of the distance in the last second. It implies that it covers first half of the distance in #(t-1)# seconds.
We obtain
#s/2=1/2g.(t-1)^2#.......(2)
Dividing equation (1) by (2)
We obtain
#2=t^2/(t-1)^2#, Solving for #t#
#2xx(t-1)^2=t^2#
or #2xx(t^2-2t+1)^2=t^2#
or #t^2-4t+2=0#
Using the quadratic formula for finding roots
#t=(-b+-sqrt(b^2-4ac))/(2a)#
or #t=(4+-sqrt((-4)^2-4xx1xx2))/2#
or #t=(4+-sqrt(8))/2#
or #t=(2+-sqrt2)#

Considering #t=(2-sqrt2)# root.

We observe that #tapprox0.6s#. For this root #(t-1)# is negative. Therefore, the condition for obtaining equation (2) can not be physically fulfilled. Hence, this root is to be ignored.
#:. t=(2+sqrt2)s#

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