Complete Question
The complete question is shown on the first uploaded image
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]