Question

Question #cdc14

Answer 1

(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`

You made a correct choice.