Ah, this is an infinite sum question, or a sum of geometric sequence
when will the sum reach 1cm of the edge or about 1.99m
so
2m=200cm
within 1cm means at least 1.99m
so
we will use m and not cm for consitancy
sum of geometric sequence is

a1=first term=initial jjump=1
r=common ratio=1/2
n=?, we ar solving for that
so
we want it to equal 1.99 so



divide both sides by 2

times -1

add 1 or 2/2 to both sides

take the ln of both sides


divide both sides by ln(1/2)

use your calculatro to find that n≈7.64386
so on 7th jump, it is not yet at 1cm to the edge but at 8th jump, it is past
so 8th jump