Question

what is this ratio of this distance covered by the freely falling object in first , second, third second of it motion​

Answer 1

$\huge \tt{ \underline{Question :}}$

→ What is this ratio of this distance covered by the freely falling object in first , second, third second of it motion?

$\huge \tt{ \underline{Answer :}}$

→ Option [4]

→ 1:3:5

$\huge \tt{ \underline{Explanation :}}$

→ The distance covered by an object during free fall is given by :

$\tt{d=ut+\dfrac{1}{2}at^2}$

→ Here, u = 0, a = g

→ So,

$\tt{d\propto t^2}$

→ For 1 second,

$\tt{d_1=kt^2=kd}$

→ For 2 second,

$\tt{d_2=k(2)^2=4kd}$

→ For 3 second,

→ Distance covered in 2 seconds,

→ 4k-k = 3k

→ Distance covered in 3 seconds,

→ 9k-4k = 5k

→ So,

→ The ratio of distances traveled by a freely falling body in first, second, and third second will be 1:3:5.

Answer 2

Step-by-step explanation:

Answer. To find, The ratio of ratio of the distances travelled by a freely falling body in first, second, and third second of its fall . So, the ratio of distances traveled by a freely falling body in first, second, and third second will be 1:3:5.

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