A body is falling freely under gravity. The distance covered by the body is first, second and fourth

Question

A body is falling freely under gravity. The distance covered by the body is first, second and fourth minutes of its motion are in the ratio\

Professor PC Thomas Classes · Answer

Here, $\mathrm{u}=0$ Distance covered in 1 minute is $S_{1}=\frac{1}{2} g(t)^{2}=\frac{g}{2}$\ Distance covered in 2 minutes is\ $S_{2}=\frac{1}{2} g(2)^{2}=\frac{4 g}{2}$\ Distance covered in 3 minutes is $$ S_{3}=\frac{1}{2} g(3)^{2}=\frac{9 g}{2} $$ Therefore, distance covered in $1^{	ext {st }}$ minute is $$ \mathrm{D}_{1}=\mathrm{S}_{1}=\mathrm{g} / 2 $$ Distance covered in $2^{	ext {nd }}$ minute is\ $\mathrm{D}_{2}=\mathrm{S}_{2}-\mathrm{S}_{1}=\frac{4 \mathrm{~g}}{2}-\frac{\mathrm{g}}{2}=\frac{3 \mathrm{~g}}{2}$\ Distance covered in 3d minute is\ $\mathrm{D}_{3}=\mathrm{S}_{3}-\mathrm{S}_{2}=\frac{9 \mathrm{~g}}{2}-\frac{4 \mathrm{~g}}{2}=\frac{5 \mathrm{~g}}{2}$\ $	herefore \mathrm{D}_{1}: \mathrm{D}_{2}: \mathrm{D}_{3}=\frac{\mathrm{g}}{2}: \frac{3 \mathrm{~g}}{2} \mathrm{~L} \frac{5 \mathrm{~g}}{2}=1: 3: 5$\ Distance travelled in the 4th minute?