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September 14, 2014

September 14, 2014

Posted by **ally** on Friday, February 25, 2011 at 8:19pm.

0.72 kg bullet is 1777 J. The acceleration of gravity is 9.81m/s2.Neglecting air resistance, ﬁnd the range of this projectile when it is ﬁred at an angle such that the range equals the maximum height attained.

Answer in units of km.

- physics -
**drwls**, Saturday, February 26, 2011 at 12:07amFirst calculate the initial velocity Vo, using

(M/2)Vo^2 = 1777 J

If the range equals the maximum height,

2(Vo^2/g)sinA*cosA = (1/2)(Vo^2/g)sin^2A

2 cosA = (1/2)sin A

tanA = 4 A = 76.0 degrees

Use that angle in either the range or the max height formula

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