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November 24, 2014

November 24, 2014

Posted by **Anonymous** on Wednesday, June 6, 2012 at 6:45pm.

- Physics -
**Elena**, Thursday, June 7, 2012 at 5:07amv(ox) =v(o) •cosα= 135 •cos15 = 130.4 m/s,

v(oy) =v(o) •sinα= 135 •sin15 = 34.9 m/s,

PlaneA:

the tank moves upward along the parabolic path until its v(y) =0.

The height hₒ= v²(o) •sin²α/2•g =135² •sin²15/2•9.8 = 62.3 m

H=h+hₒ=2000+62.3 =2062.3 m.

t =sqrt(2H/g) = sqrt(2•2062.3/9.8) = 20.5.

The vertical component of the velocity near the ground is

v(y)=gt = 9.8•20.5 = 201.05 m/s.

v =sqrt(v²(x)+v²(y)) = sqrt(130.4² +201.05²) = 239.6 m/s.

tanφ =v(y)/v(x) =201.05/130.4 =1.54,

φ=57º

Plane B.

h=gt²/2 => t=sqrt(2h/g) =sqrt(2•2000/9.8) = 20.2 s.

v(y) = v(oy)+gt = 34.9+9.8•20.2 = 232.9 m/s.

v =sqrt(v²(x)+v²(y)) = sqrt(130.4² +232.9²) = 266.9 m/s.

tanφ =232.9/130.4 = 1.789,

φ = 60.8º.

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