Question

9, You are piloting a helicopter which is man vertically at a un

you reach 196.00 m,

velocity of 8.50 m/

a) When does th

14.70 m/s. When

a horizontal

B vertically ata uniform velocity of 14.70 m/s

ach 196.00 m, you see Barney (Uh-oh). A la

1. A large object is projected with a horizon

city of 8.50 m/s from the rising helicopter.

Vhen does the ball reach Barney's head if he

ad if he is standing in a hole with his head at

level? (7.99 s]

b) Where do

have to be horizontally relative to the helicopter's position? (68.0 ml

city when it hits the ground? - 63.7 m/s]

c) What is the vertical velocity when it hits the

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   t =  8 seconds

b

  $D = 68 \ m$

c

  $v_y = 63.7 \ m/s$

Explanation:

From the question we are told that

   The initial uniform velocity is $v =14.70 \ m/s$

    The height considered is  $s = 196.00 \ m$

   The speed of the large object is  $v = 8.50 \ m/s$

Generally according to  the equation of motion

    $s = -ut + \frac{1}{2} gt^2$

Here  u is  negative because the helicopter is moving upward against gravity

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

=> $4.9 t^2 - 14.7t -196 = 0$

solving this using quadratic equation formula we obtain that

 t =  8 seconds

So the time taken for the large object to hit Barney is  t =  8 seconds

Gnerally the horizontal distance of Barney relative to the helicopter is mathematically represented as

      $D = v * t$

       $D = 8.5 * 8$

     $D = 68 \ m$

generally given that Barney head is the same level with the ground then the vertical velocity when it hits the ground is mathematically represented as

     $v_y = -u +gt$

=>   $v_y = -14.7 +9.8 * 8$

=>   $v_y = 63.7 \ m/s$