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March 5, 2015

March 5, 2015

Posted by **Mima** on Sunday, September 16, 2012 at 4:52pm.

- physics -
**Elena**, Monday, September 17, 2012 at 4:34pmv₀⒳= v₀•cosα,

v₀⒴ =v₀•sinα,

v₀² =v₀⒳² +v₀⒴²,

v² =v⒳² +v⒴²,

v⒳= v₀⒳,

v⒴=v₀⒴ -g•t,

v² =v⒳² +v⒴² = v₀⒳²+(v₀⒴ -g•t)²=

=v₀⒳²+v₀⒴² -2 v₀⒴•g•t +(g•t)² =

= v₀²-2 v₀•sinα•g•t +(g•t)².

sinα= [v₀²-v²+(g•t)²]/2•v₀•g•t =

=[14.5² -12.4²+(9.8•0.272²]/(2•14.5•9.8•0.272) =

=0.823.

α=55.35°

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