Consider two identical baseballs which start at the same location; their terminal speed at this location is 32 m/s. At t = 0s, Ball A is thrown straight up at m/s; Ball B is thrown at 16m/s 30.0° below the horizontal at 32 m/s. How do the magnitudes of the baseballs' accelerations32 2at t = 0s compare (do not neglect air resistance)?

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Complete Question

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Answer:

The correct option is  B

Explanation:

From the question we are told that

  The terminal  velocity of both balls is  [tex]v_t  =  32 \ m/s[/tex]

   The speed with which ball A is thrown up at t =0 s is  [tex]v__A}}  =  16\sqrt{2}[/tex]

   The speed with which ball B is thrown 30° below the horizontal  at t =0 s is  

[tex]v__{B}} =  32\sqrt{2}[/tex]

Generally the acceleration of ball A is (g i.e acceleration due gravity)given that it is thrown straight up  

Generally the vertical  acceleration of ball B is mathematically represented as

    [tex]a  = gcos (90 - 30)[/tex]

      [tex]a  = gcos (60)[/tex]

Generally the ratio of both acceleration is  

     [tex]\frac{a__{{A}}}{a__{{B}}}} =\frac{g}{gsin(60)}[/tex]

=> [tex]\frac{a__{{A}}}{a__{{B}}}} = \frac{1}{sin (60)}[/tex]

=>  [tex]\frac{a__{{A}}}{a__{{B}}}} = \frac{1}{\frac{\sqrt{3} }{2} }[/tex]

=>[tex]\frac{a__{{A}}}{a__{{B}}}} = \frac{2}{\sqrt{3} }[/tex]

Rationalizing

=> [tex]\frac{a__{{A}}}{a__{{B}}}} = \frac{2}{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} }[/tex]

=> [tex]\frac{a__{{A}}}{a__{{B}}}} = \frac{2 \sqrt{3} }{3}[/tex]

=>[tex]a__{B}} = \frac{2\sqrt{3} }{3} * a__{A}}[/tex]

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