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The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second :  
  • a)
    1 :  4 :  9 :  16
  • b)
    1 : 3 : 5 : 7  
  • c)
    1 :  1 :  1 :  1
  • d)
    1 : 2 : 3 : 4 
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The ratio of the distances travelled by a freely falling body in the 1...
< b="" />Explanation: < />

The distance travelled by a freely falling body in each second is given by the formula d = (1/2)gt^2, where d is the distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time in seconds.

To find the ratio of the distances travelled in the 1st, 2nd, 3rd, and 4th second, we can substitute the values of t and calculate the distances.

< b="" />1st Second:< />
d1 = (1/2)g(1)^2 = (1/2)g

< b="" />2nd Second:< />
d2 = (1/2)g(2)^2 = 2g

< b="" />3rd Second:< />
d3 = (1/2)g(3)^2 = (9/2)g

< b="" />4th Second:< />
d4 = (1/2)g(4)^2 = 8g

Now, let's calculate the ratio of the distances:

< b="" />Ratio:< />
d1 : d2 : d3 : d4
= (1/2)g : 2g : (9/2)g : 8g

To simplify the ratio, we can divide each term by (1/2)g:

< b="" />Simplified Ratio:< />
1 : 4 : 9 : 16

Therefore, the correct answer is option < b="" />'A'< />, which states that the ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd, and 4th second is 1 : 4 : 9 : 16.
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Community Answer
The ratio of the distances travelled by a freely falling body in the 1...

= [2(1) – 1] : [2(2) – 1] : [2(3) – 1] : [2(4) – 1] 
 = 1 :  3 :  5 :  7
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The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second : a)1 : 4 : 9 : 16b)1 : 3 : 5 : 7c)1 : 1 : 1 : 1d)1 : 2 : 3 : 4Correct answer is option 'B'. Can you explain this answer?
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