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April 18, 2014

April 18, 2014

Posted by **Tracie** on Wednesday, September 5, 2012 at 12:55am.

- physics -
**Henry**, Thursday, September 6, 2012 at 5:27pmd1 + d2 = 5.97 m.

Vo1*t+o.5g*t^2 + Vo2*t+0.5g*t^2 = 5.97

8.5t + 4.9t^2 + 8.5t - 4.9t^2 = 5.97

17t = 5.97

t = 0.351 s. = Time at which they met.

h1 =5.97-(8.5*0.351 + 4.9*(0.3510^2) = 2.38 m. Above base of cliff.

h2 = 8.5*0.351 - 4.9(0.351)^2 = 2.38 m.

Above the base of cliff.

The stones cross paths at 2.38 m above base of cliff.

- physics -
**ailabianca**, Friday, January 10, 2014 at 7:34pmthe other straight downward from the top of the cliff. The stones are thrown with the same speed. The height of the cliff is 6.00 m, and the speed with which the stones are thrown is 9.00 m/s. Find the location of the crossing point.

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