The distances covered by a freely falling body in its first, second, third, ..., nth seconds of its motion

Question

The distances covered by a freely falling body in its first, second, third, ..., $n^{th}$ seconds of its motion

Smart Achievers · Accepted Answer

(1)

Distance travelled by a body in $n^{th}$ second is

$D_{n} = u + \frac{a}{2} (2 n - 1)$

Here, $u = 0, a = g$

$∴$ Distance travelled by the body in $1^{st}$ second is

$D_{1} = 0 + \frac{g}{2} (2 \times 1 - 1) = \frac{1}{2} g$

Distance travelled by the body in $2^{nd}$ second is

$D_{2} = 0 + \frac{g}{2} (2 \times 2 - 1) = \frac{3}{2} g$

Distance travelled by the body in $3^{rd}$ second is

$D_{3} = 0 + \frac{g}{2} (2 \times 3 - 1) = \frac{5}{2} g$ and so on.

Hence, the distances covered by a freely falling body in its first, second, third, ......, $n^{th}$ second of its motion forms an arithmetic progression.

Solution

Q.
The distances covered by a freely falling body in its first, second, third, ..., $n^{th}$ seconds of its motion

1 forms an arithmetic progression

2 forms a geometric progression

3 do not form any well defined series

4 form a series corresponding to the difference of square root of the successive natural numbers

Want Us to Email you About Special Offers & Updates?