Let x be the height in cm from the ground then x=asin(b(t+p))+c, where p is the phase difference, and a, b, c are constants. At t=3, x=40 and sin(b(t+p))=-1 (minimum) so 40=c-a, c=a+40, and (3+p)b=3(pi)/2, so b=3(pi)/(2(3+p). Now we have x=asin(3(pi)(t+p)/2(3+p))+a+40. Also, the length of the string, y, is 62cm when x=40 (because the yo-yo is fully extended). The yo-yo is suspended from a point 62+40=102cm from the ground.
At t=0, x=asin(3p(pi)/(2(3+p))+a+40=102; a(sin(3p(pi)/(2(3+p))+1)=62.
When t=2+3=5 the yo-yo is back at its lowest position again: 3(pi)(5+p)/(2(3+p)=7(pi)/2.
3(5+p)=7(3+p); 15+3p=21+7p; 4p=-6, p=-3/2 and sin(3p(pi)/(2(3+p))=sin(-3(pi)/2)=1
So a=62/(1+1)=31 and x=31sin((pi)(t-3/2))+71.
At t=20 (10 cycles), x=102cm. 20 seconds is equivalent to 10 cycles of 2 second periods, so at 20 seconds the yo-yo is in the same position as it was at the start, in other words the length of the string plus the minimum height from the ground, 102cm.
When x=52, 52=31sin((pi)(t-3/2))+71, sin((pi)(t-3/2))=-19/31; (pi)(t-3/2)=-0.6597; t=1.29 seconds.
However, since the cycle time is 2 seconds, there must be a minimum at 1 second (x=40cm) and therefore there must be another value of t less than 1 second, when x=52cm, because the yo-yo must pass the 52cm height to get to 40cm. The sine curve is symmetrical about each minimum, so, if x=52cm at 1.29 seconds, it must also be 52cm at 1-0.29=0.71 seconds. Therefore the earliest time is t=0.71 seconds.