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January 28, 2015

January 28, 2015

Posted by **Jacqueline** on Friday, December 2, 2011 at 12:22am.

- Physics Honors -
**Steve**, Friday, December 2, 2011 at 12:31pmthe equation of motion for an object thrown from (0,0) at an angle θ with velocity v is

y(x) = -g/(2v^2 cos^2 θ) x^2 + xtanθ

the range (where y=0 again) is

r = v^2 sin2θ/g

the maximum height reached is

h = v^2 sin^2 θ/2g

So, we have

θ = 60°

v = 25

r = 25^2 * sin 120°/9.8

= 625 * 0.866 / 9.8

= 55.2m

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