Solution 1The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:\\\\(v^2=u^2+2aS\\\\), where \\\\(u=0\\\\)The ball with higher speed will take less time to cross the window.\\\\(\\because\\\\) \\\\(H_2 > H_1\\\\) \\\\(\\implies\\\\) \\\\(v_2 > v_1\\\\) \\\\(\ herefore\\\\) \\\\(t_2 < t_1\\\\)Hence Option D is correctSolution 2Statement D is correctSituation 1:We want time to cross x.\\\\(v^{2}=u^{2}+2as\\\\)\\\\(u=0, a=-g, s=-1 \\\\)\\\\(v^{2}=2g\\Rightarrow v_{1}=\\sqrt{2g}\\\\)\\\\((1+x)=\\dfrac{1}{2} gt^{2}\\RightarrowT_{2}=\\sqrt{\\dfrac{2(1+x)}{g}}\\\\)\\\\(1=\\dfrac{1}{2}gt^{2}=T_{1}=\\sqrt{\\dfrac{2}{g}}\\\\)So time taken to cross \\\\(x=t_{1}=T_{2}-T_{1}=\\sqrt{\\dfrac{2}{g}}(\\sqrt{1+x}-1)\\\\)Situation 2:\\\\((2+x)=\\dfrac{1}{2}gt^{2}\\RightarrowT_{2}=\\sqrt{\\dfrac{2(2+x)}{g}}\\\\)\\\\(2=\\dfrac{1}{2}gt^{2}\\Rightarrow T_{1}=\\sqrt{\\dfrac{2\ imes 2}{g}}\\\\)So time taken to cross \\\\(x=t_{2}=T_{2}-T_{1}=\\sqrt{\\dfrac{2}{g}}(\\sqrt{2+x}-\\sqrt{2})\\\\)Clearly \\\\(t_{2}< t_{1} \\\\)