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April 18, 2014

April 18, 2014

Posted by **Joey** on Thursday, September 16, 2010 at 4:35pm.

- physics -
**bobpursley**, Thursday, September 16, 2010 at 6:35pmh=1/2 g t^2 is the basic formula.

H=1/2 g (t-to)^2 where to is the time the lower one is dropped, after the higher.

4H=1/2 g t^2

So, both balls hit at time t.

t=sqrt 8H/g

H=1/2 g (sqrt(8H/g)-to)^2

2H=g(8H/g-2to*sqrt(8H/g)+to^2)

so you have a quadratic...

to^2-2to* sqrt(8H/g)-2H/g

a=1 b=-2sqrt(8H/g) c= -2H/g

solve by the quadratic equation.

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