Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 63: 66

The difference in time taken is $0.1s.$

At Petronas Twin Towers in Malaysia, let time taken for drop = $t_{1}$ $S=ut+\frac{1}{2}at^{2}=0+0.5\times 9.8t_{1}^{2}=452$ Or, $t_{1}=9.6s$ At Chicago Sears Tower, let time taken for drop = $t_{2}$ $S=ut+\frac{1}{2}at^{2}=0+0.5\times 9.8t_{2}^{2}=443$ Or, $t_{2}=9.5s$ The difference in time taken = $t_{1}-t_{2}=9.6-9.5=0.1s$