Answer
To prevent pancake collapse of the building, the safety factor $s$ must exceed the value of $~~602$
Work Step by Step
We can find the speed after falling a distance of $4.0~m$:
$v_f^2 = v_0^2+2ay$
$v_f^2 = 0+2ay$
$v_f = \sqrt{2ay}$
$v_f = \sqrt{(2)(9.8~m/s^2)(4.0~m)}$
$v_f = 8.854~m/s$
We can find an expression for the momentum of the falling weight $W$ after falling 4.0 meters:
$p = m~v_f$
$p = \frac{W~v_f}{g}$
In order to oppose the momentum of the falling weight, there must be an impulse from the lower floor exerted upward on the falling weight. We can find the magnitude of the required impulse:
$J = p$
$J = \frac{W~v_f}{g}$
We can find the force that is required during the time of the collision:
$F~t = J$
$F = \frac{J}{t}$
$F = \frac{W~v_f}{g~t}$
$F = \frac{v_f}{g~t}~W$
$F = \frac{8.854~m/s}{(9.8~m/s^2)(1.5\times 10^{-3}~s)}~W$
$F = 602~W$
Therefore, to prevent pancake collapse of the building, the safety factor $s$ must exceed the value of $~~602$
