Answer:
[tex] t2/t1 = \sqrt{2} - 1 [/tex]
Explanation:
The expression for the second law of motion is given below:
h = [tex] ut + 0.5at^2 [/tex]
For first half distance
Object is initially at rest, so its initial speed u = 0
Object falls at half the distance, so h = h/2 where t = t1
Hence, we have
[tex] h/2 = at1^2/2 - equation 1[/tex]
For second half distance:
Similarly,
[tex] h = a(t1 + t2)^2/2 - equation 2 [/tex] where t = t1 + t2 and u= 0
Using equation 2 by equation 1
we obtain [tex] 2 = (t1 + t2)^2/t1^2 [/tex]
Hence [tex] t2/t1 + 1 = \sqrt{2} [/tex]
Hence [tex] t2/t1 = \sqrt{2} - 1 [/tex]
