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October 20, 2014

October 20, 2014

Posted by **prve** on Wednesday, October 22, 2008 at 3:53pm.

- physics -
**GK**, Wednesday, October 22, 2008 at 4:48pmI 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]

- physics -
**prve**, Wednesday, October 22, 2008 at 5:25pmCould you simplify your answer into simple words

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