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Find  the  ratio  of  the  distance  travelled  by  a freely  falling  body  during  the  first  and  third second of its fall.    (a) 1 : 3  (b)  3 : 5             (c)  1 : 5  (d)  2 : 3

Question

Find  the  ratio  of  the  distance  travelled  by  a freely  falling  body  during  the  first  and  third second of its fall.    (a) 1 : 3  (b)  3 : 5             (c)  1 : 5  (d)  2 : 3

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Solution

The distance travelled by a freely falling body under gravity in nth second can be given by the equation:

d = u + 1/2 * g * (2n - 1)

where:

  • d is the distance travelled in the nth second,
  • u is the initial velocity (which is 0 for a freely falling body),
  • g is the acceleration due to gravity, and
  • n is the nth second.

For the first second (n=1), the distance travelled is:

d1 = 0 + 1/2 * g * (2*1 - 1) = 1/2 * g

For the third second (n=3), the distance travelled is:

d3 = 0 + 1/2 * g * (2*3 - 1) = 5/2 * g

The ratio of the distance travelled during the first and third second is therefore:

d1 : d3 = (1/2 * g) : (5/2 * g) = 1 : 5

So, the correct answer is (c) 1 : 5.

This problem has been solved

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1/3

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