Question

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?

Answer 1

Answer:

The ball reaches Barney  head in  $t = 8 \ s$

Explanation:

From the question we are told that

 The rise velocity is  $v = 14.70 \ m/s$

  The height considered is $h = 196 \ m$

   The horizontal velocity of the large object is  $v_h = 8.50 \ m/s$

   

Generally from kinematic equation  

   $s = ut + \frac{1}{2} gt^2$

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  

     $u = -14.7 m/s$

So

    $196 = -14.7 t + \frac{1}{2} * 9.8 * t^2$

=  $4.9 t^2 - 14.7t - 196 = 0$

Solving the above equation using quadratic formula  

    The value of  t obtained is  $t = 8 \ s$