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