Question

What will be the distance moved by a freely falling body in nth second of its motion?(initial velocity=0)

Answer 1

Answer on Question #48376, Physics, Mechanics - Kinematics - Dynamics

What will be the distance moved by a freely falling body in nth second of its motion? (initial velocity=0)

Distance covered by the falling object is equal to:

$$S = \frac {g t ^ {2}}{2}$$

Where $t$ is a number of seconds.

Therefore, moved distance in nth second:

$$S _ {1} = \frac {g}{2}$$ $$S _ {2} = 4 \frac {g}{2} - \frac {g}{2} = 3 \frac {g}{2}$$ $$S _ {3} = 9 \frac {g}{2} - 4 \frac {g}{2} = 5 \frac {g}{2}$$ $$S _ {n} = (n ^ {2} - (n - 1) ^ {2}) \frac {g}{2} = (2 n - 1) \frac {g}{2}$$

Answer: $S_{n} = (2n - 1)\frac{g}{2}$

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