Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 64: 73

1.48m

Let height above top of the window=h. $S=ut+\frac{1}{2}at^{2}=0+\frac{1}{2}gt^{2}$ Therefore, $S=4.9t^{2}$ $S_{1}=4.9t_{1}^{2}$ $S_{2}=4.9t_{2}^{2}$ So, $S_{1}-S_{2}=4.9 \times (t_{1}-t_{2})(t_{1}+t_{2})$ i.e. $1.35=4.9\times 0.21\times (t_{1}+t_{2})$ $(t_{1}+t_{2})=1.31s$ Therefore, $t_{1}=0.76s$ and $t_{2}=0.55s$ So, $S_{2}=4.9t_{2}^{2}=4.9\times 0.55^{2}=1.48m$