The ratio of the displacements of a freely falling body during first, second and third seconds of its motion is

Distance travelled in $\mathrm{n}^{\text {th }}$ second is given by $$ \mathrm{S}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) $$ When a body is free falling $\mathrm{a}=\mathrm{g}$ Now, Distance travelled in first second : $(n=1)$ $$ \mathrm{S}_1=0+\frac{\mathrm{g}}{2}(2-1)=\frac{\mathrm{g}}{2} $$ Distance travelled in 2nd second: $(n=2)$ $$ \mathrm{S}_2=0+\frac{\mathrm{g}}{2}(2 \times 2-1)=\frac{3 \mathrm{~g}}{2} $$ Distance travelled in 3rd second : $(n=3)$ $$ \mathrm{S}_3=0+\frac{\mathrm{g}}{2}(2 \times 3-1)=\frac{5 \mathrm{~g}}{2} $$ Hence, the ratio is given by: $$ \begin{aligned} & \mathrm{S}_1: \mathrm{S}_2: \mathrm{S}_3=\frac{\mathrm{g}}{2}: \frac{3 \mathrm{~g}}{2}: \frac{5 \mathrm{~g}}{2} \\ & =1: 3: 5 \end{aligned} $$