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You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Barney (Uh-oh). A large object is projected with a horizontal velocity of 8.50 m/s from the rising helicopter. When does the ball reach Barney's head if he is standing in a hole with his head at ground level?

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okpalawalter8

Answer:

The ball reaches Barney  head in  [tex]t = 8 \ s[/tex]

Explanation:

From the question we are told that

 The rise velocity is  [tex]v = 14.70 \ m/s[/tex]

  The height considered is [tex]h = 196 \ m[/tex]

   The horizontal velocity of the large object is  [tex]v_h = 8.50 \ m/s[/tex]

   

Generally from kinematic equation  

   [tex]s = ut + \frac{1}{2} gt^2[/tex]

Here s is the distance of the object from Barney head ,

        u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So  

     [tex]u = -14.7 m/s[/tex]

So

    [tex]196 = -14.7 t + \frac{1}{2} * 9.8 * t^2[/tex]

=  [tex]4.9 t^2 - 14.7t - 196 = 0[/tex]

Solving the above equation using quadratic formula  

    The value of  t obtained is  [tex]t = 8 \ s[/tex]

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