Distance travelled by a body during nth second of its fall

• Derive the formula for distance covered by a body during nth second of its fall
• Explain special cases of the obtained formula

Distance travelled by a body during n^th second of its fall denotes through what distance the position of the body has changed in a particular time of fall. If a body reaches the surface of the Earth in 5 seconds from a height of 122.5 m then, the distance it covers in 1st second, 2nd second, 3rd second, 4th second, and 5th second of its fall is regarded as the distance travelled by that body during nth second of its fall. If the body had covered covered 78.4m in 4 seconds while it had covered 50m in 5 seconds, then the space covered by the body in the 5th second of its fall is obtained by subtracting the distance travelled by it in 5 seconds and that in 4 seconds. Hence, the body had travelled 44.1m in the 5th second of its fall. It can be further generalized as: distance covered by a body in n^th second is equal to the difference of distance travelled in n seconds and distance travelled in (n-1) seconds.

Derivation of an Expression

Let us consider a body of a given mass is dropped from a height ‘h’ under a constant downward acceleration equal to acceleration due to gravity ‘g’ reaches the surface of the Earth in time ‘t’. Then, u = 0, s = h and a = g. Let $\text{s}_\text{t}$ be the distance travelled by the body in t seconds.

From equations of laws of motion, the height travelled by the body in its total time of fall is given by

$$\text{s = ut + }\frac{1}{2} \text{gt}^2$$

If t = n, then in n seconds, the body covers distance equal to

$$\text{s}_\text{n} \text{ = un + }\frac{1}{2} \text{gn}^2\text{ — (i)}$$

If t = n – 1, then in (n-1) seconds, the body covers distance equal to

$$\text{s}_\text{n-1} \text{ = u(n-1) + }\frac{1}{2} \text{g(n-1)}^2 \text{ — (ii)}$$

Following our generalization mentioned in the introduction, the distance covered by the body in nth second of its fall would be

$$\text{s}_\text{n} \text{ – s}_\text{n-1}\text{ = un + }\frac{1}{2} \text{gn}^2 \text{ – } \left ( \text{u(n-1) + }\frac{1}{2} \text{g(n-1)}^2 \right )$$

$$\text{= u(n – (n – 1)) +} \frac{1}{2} \text{g} \left ( \text{n}^2 \text{ – (n – 1)}^2 \right ) $$

$$\text{= u(n – n + 1) +} \frac{1}{2} \text{g} \left ( \text{n}^2 \text{ – n}^2 \text{ + 2n – 1} \right )$$

$$\text{ – s}_\text{n-1} \text{ = u(1) +} \frac{1}{2} \text{g (2n – 1)}$$

Hence, the distance covered by a body in nth second of its fall is given by

$$\text{s}_\text{n} \text{ – s}_\text{n-1} \text{ = u +} \frac{1}{2} \text{g (2n – 1)}$$

Special Cases

• When the body has no initial velocity, eg: condition of free fall or starting from rest, then u = 0. Thus, our formula reduces to $$\text{s}_\text{n} \text{ – s}_\text{n-1} \text{ = }\frac{1}{2} \text{g (2n – 1)}$$
• When the body has no net acceleration, then g = 0. Thus, our formula reduces to $$\text{s}_\text{n} \text{ – s}_\text{n-1} \text{ = u}$$This suggests that the body covers equal distance in equal interval of time under zero acceleration but a initial velocity. This also suggests that the velocity of the body is constant throughout the motion.
• This formula can be applied to any bodies having a constant acceleration. When a body is traveling displacement in horizontal direction then, the acceleration due to gravity is replaced by the acceleration of the body. Thus, the modified formula would be: $$\text{s}_\text{n} \text{ – s}_\text{n-1} \text{ = u +} \frac{1}{2} \text{a (2n – 1)}$$