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

Complete Question

The complete question is shown on the first uploaded image

a

   t =  8 seconds

b

c

Explanation:

From the question we are told that

   The initial uniform velocity is

    The height considered is  

   The speed of the large object is  

Generally according to  the equation of motion

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

=>

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

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

=>  

=>  

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.