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

Two planes are about to drop an empty fuel tank. At the moment of release each plane has the same speed of 135 m/s, and each tank is at the same height of around 2.00km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15deg above the horizontal (Plane A) and the other is flying at an angle of 15deg below the horizontal (Plane B). Find the magnitude and the direction of the velocity with which the fuel tank hits the ground if it is from a) Plane A b) Plane B. Give the directional angles with respect to the horizontal.

- Physics -
**Elena**, Thursday, June 7, 2012 at 5:07am
v(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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