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March 28, 2015

March 28, 2015

Posted by **laura** on Wednesday, September 14, 2011 at 6:45pm.

- Tulane Physics -
**bobpursley**, Wednesday, September 14, 2011 at 8:09pmconsider the vertical movement

hf=hi+Viv*t-1/2 g t^2

first equation: 0=Viv*3.56-4.9(3.56^2)

solve for Viv, the initial componentof vertical velocity

Second equation: hf=hi+Viv*5.31-4.9(5.31)^2

0=hi+ Viv....

solve for hi, the initial height of the cliff.

Finally, initial horizontal velocity (at 45 deg) must be the same magnitude as the initial vertical component.

horizontal distance= Vih*5.31

where Vih=Viv

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