Posted by abebe on Friday, June 15, 2012 at 11:45am.
Without air RESISTANCE(!), the horizontal component of the velocity is constant
v(x) =v(ox) = v•cos30 = 20•cos30 = 17.32 m/s.
The path of the projectile is symmetrical respectively the max height, therefore , there is the point("left" point) on the ascending branch of parabolic trajectory at the height h= 10 m.
The magnitude of the vertical component of velocity v(y) at the “left” point is equal to the magnitude v(y) at the “right” point (their directions are opposite). Vertical component of the initial velocity and vertical component of velocity at height 10 m are related as
v(oy)² - v(y)² = 2•g•h, =>
v(oy) =sqrt {2•g•h+v(y)²} =
= sqrt {2•g•h+(v•sin30)²} =
= sqrt {2•9.8•10+(20•sin30)²} =17.2 m/s.
v(o) =sqrt{v(ox)²+v(oy)²} =
= sqrt{17.32²+ 17.2²} = 24.4 m/s
tan α =v(oy)/v(ox) = 17.2/17.32=0.99
α =44.8º
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