Question

What is the ratio of the distance traveled by a body falling freely from rest in the first, second, and third seconds of its fall?

Answer 1

Step 1:Given data;

Initial velocity of the body = u = 0 m/s

Acceleration due to gravity = g m/s²

Step 2: Formula used:

We have the 2nd equation of motion,

$S=ut+\frac{1}{2}a{t}^2$

Where s is distance traveled, u is initial velocity, a is acceleration, and t is time.

Step 3:Finding the ratio:

$$\begin{gathered} \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{Distance}\mathrm{covered}\mathrm{in}{\mathrm{n}}^{\mathrm{th}}\mathrm{second}; \\ {\mathrm{S}}_{{\mathrm{n}}^{\mathrm{th}}}=\mathrm{u}+\frac{1}{2}\mathrm{g}(2\mathrm{n}-1) \\ {\mathrm{S}}_1=\frac{\mathrm{g}}{2}\times (2\times \left(1\right)-1)=\frac{\mathrm{g}}{2} \\ {\mathrm{S}}_2=\frac{\mathrm{g}}{2}\times (2\times \left(2\right)-1)=\frac{3\mathrm{g}}{2} \\ {\mathrm{S}}_3=\frac{\mathrm{g}}{2}\times (2\times \left(3\right)-1)=\frac{5\mathrm{g}}{2} \\ {\mathrm{S}}_1:{\mathrm{S}}_2:{\mathrm{S}}_3=\frac{\mathrm{g}}{2}:\frac{3\mathrm{g}}{2}:\frac{5\mathrm{g}}{2}=1:3:5 \end{gathered}$$

Thus, the ratio of distance traveled by a body falling freely from rest in the first, second, and third seconds of its fall is 1:3:5.