Let ss be the total distance covered in time tt.
We know from general expression that s=ut+1/2g.t^2s=ut+12g.t2
Since the object falls from rest, u=0u=0
Equation becomes s=1/2g.t^2s=12g.t2 ......(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)(t−1) seconds.
We obtain
s/2=1/2g.(t-1)^2s2=12g.(t−1)2.......(2)
Dividing equation (1) by (2)
We obtain
2=t^2/(t-1)^22=t2(t−1)2, Solving for tt
2xx(t-1)^2=t^22×(t−1)2=t2
or 2xx(t^2-2t+1)^2=t^22×(t2−2t+1)2=t2
or t^2-4t+2=0t2−4t+2=0
Using the quadratic formula for finding roots
t=(-b+-sqrt(b^2-4ac))/(2a)t=−b±√b2−4ac2a
or t=(4+-sqrt((-4)^2-4xx1xx2))/2t=4±√(−4)2−4×1×22
or t=(4+-sqrt(8))/2t=4±√82
or t=(2+-sqrt2)t=(2±√2)
Considering t=(2-sqrt2)t=(2−√2) root.
We observe that tapprox0.6st≈0.6s. For this root (t-1)(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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