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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of explorers. The acceleration of gravity is 9.8 m/s 2 . Where does the package strike the ground relative to the point directly below where it was released? Answer in units of m. 008 (part 2 of 3) 10.0 points What is the horizontal component of the velocity just before it hits? Answer in units of m/s. 009 (part 3 of 3) 10.0 points What is the vertical component of the velocity just before it hits? (Choose upward as the positive vertical direction) Answer in units of m/s.

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Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

onyebuchinnaji

The horizontal distance traveled by the package below the point it was released is 256.66 m.

The horizontal component of the velocity before it hits the ground is 41 m/s.

The vertical component of the velocity before it hits the ground is -61.35 m/s.

The given parameters;

  • initial velocity of the plane, u = 41 m/s
  • height of the plane above the ground, h = 192 m

The time of motion of the package before it strikes the ground is calculated as;

[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 192}{9.8} }\\\\t = 6.26 \ s[/tex]

The horizontal distance traveled by the package below the point it was released;

X = ut

X = 41 x 6.26

X = 256.66 m

The horizontal component of the velocity before it hits the ground is the same as the initial since horizontal velocity is not affected by gravity.

[tex]v_x_f = v_x_0 = 41 \ m/s[/tex]

The vertical component of the velocity before it hits the ground is calculated as;

[tex]v_f_y = v_0_y - gt\\\\v_f_y = 0 - 9.8 \times 6.26\\\\v_f_y = -61.35 \ m/s[/tex]

Learn more here:https://brainly.com/question/20349275

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