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
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.
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