A stone is dropped from the top of a cliff and is found to travel 44.1m in the last second before it reaches the ground. Find the height of the cliff. (Can we use 1:3:5 method if yes then how)

we know
distance travelled in the nth second Sn= u+ 1/2*a* (2n-1)

here
Sn=44.1
u=0 (initial velocity)
a=9.8 m/s^2
since the time we taken is the last second n=t
where t is the time taken to reach the ground

Sn= u+ 1/2*a* (2n-1)
44.1 = 0 + 0.5*9.8*(2t-1)
44.1 = 4.9 (2t-1)
44.1 = 9.8t -4.9
44.1+4.9 = 9.8t
9.8t = 49
t =49/9.8 =5 sec.

So the ball takes 5 seconds to reach the ground.
Now by using the formula
S=ut + 1/2*a*t^2

height of the cliff is = (0*5)+(0.5*9.8*5*5)
=0+0.5*9.8*5*5
=122.5 m.

You don't want to use the 1:3:5:...ration since we doesn't have data about in which second
it travelled 44.1m.