How many secons apart should two balls be released so that a ball being deopped from a height of H reaches the ground at the same time as a ball dropped from a height of 5H?

Answer 1

I do not believe the question can be answered with a specific value. Here are my reasons:
H = gt^2/2
t = sqrt[2H/g]
Let the time for 5H = t'
5H = gt'^2/2
t' = sqrt[10H/g] = sqrt(5)*sqrt[2H/g]
The ratio of the two times is:
t'/t = sqrt(5)*sqrt[2H/g] / sqrt[2H/g] = sqrt(5)
or
t' = t*sqrt(5)
The 2nd ball must be released ( t' - t ) seconds later. The actual number of seconds depends on the value of H:
( t' - t ) = sqrt(5)*sqrt[2H/g] - sqrt[2H/g]
or
( t' - t ) = sqrt[2H/g]*[sqrt(5)-1]

Answer 2

Could you simplify your answer into simple words