Question

A rooster escapes from the farm. The rooster propels itself off the roof of a 160 foot tall barn at an angle of 40° and has an initial velocity of 15 m/s. After traveling for 30 seconds at a constant speed, the rooster decides to lower down by completely stopping all flying motion. After free falling 10 meters, an eagle traveling directly leftward hits the rooster. The eagle was flying at a constant 45 m / s.

a) Draw a diagram depicting the situation, being sure to include trajectory paths.

b) Assuming the rooster gets knocked unconscious by the eagle and falls to the ground, calculate how far from the barn the rooster ends up. Neglect air resistance.

c) A veterinarian claims that the rooster wouldn’t have been knocked unconscious if the eagle was traveling at 30 m / s. If this was the case, and assuming a reaction time of 1.5 seconds after the impact, how far would the rooster fall before beginning to regain flight? Neglect air resistance.

Edit #1: Professor said by "stopping all flying motion" it means the rooster stopped flapping its wings. Its velocity began to slow down at this point because the rooster was no longer putting in any work to maintain the velocity it was previously traveling at.

Answer 1

Firstly certain things should be kept in mind while solving such questions, break the questions into small parts and then solve it. Always work in SI units.

a)

b) First we'll find how far was the rooster from the barn when it collided with the eagle.

1. Constant velocity(15m/s) flight at 40deg for 30s.

The rooster flew from the top of the barn with a constant speed of 15m/s at an angle of 40deg, for 30 sec.

distance travelled by it = 15*30 = 450 m.

horizontal distance travelled = 450cos(40) = 344.72m

2. Flying motion stops.

The rooster will keep going up until its vertical velocity component = 0.

time taken until the u_y = 0, t = usin(40)/g                      (v = u + at)

t = 15sin(40)/10 = 0.964s.

horizontal distance travelled by the rooster in time t = 15cos(40)*0.964 = 11.077 m.

After time t, rooster free falls and collides from the eagle.

Thus at the time of collision rooster was 355.8m away from the barn.

3. At what height did the rooster enter free fall.

during contant velocity flight rooster attained a height = 450sin(40) = 30m (above the barn)

After it stopped the flying height attained by it = (usin(40))²/2g               (v² - u² = 2as)

                                                                  = (9.64)²/20 = 4.64m

or 30 + 4.64 = 34.64 m (above the barn)

Total height above the ground = height of barn + height attained above the barn = 48.77 + 34.64 = 83.42m

Thus the collision occured at a height of h = 73.42 m above the ground level.

4. After collision

As no external force is acting in the horizontal direction, momentum will stay conserved in this direction

Initial momentum = 45M             M is the mass of the eagle.

Applying conservation of Momentum

45M = Mv + Mv

**(Assuming eagle and rooster have the same mass M and they move with same velocities post collision )

v_x = 22.5m/s

Vertical velocity of the rooster remains the same = v_y = 10²/2g = 5m/s                  (s = ut + at²/2)

time taken by the rooster to reach the ground after collision

w² = v_y² + 2gh ( w = velocity just before reaching the ground, h = 73.42m)

w² = 5² + 20*73.42.

w = 38.64m/s

w = v_y + gt'

t' = (38.64 - 5)/10 = 3.364s.

Horizontal distance travelled by rooster in time t' = 22.5*3.364 = 75.69m.

Therefore the rooster hits the ground at a distance of 355.58 - 75.69 = 279.89m away from the barn.

c) If the eagle was flying at a contant velocity of 30m/s, after collision horizontal velocity of the rooater would be

v_x = 15m/s        (conservation of momentum in horizontal direction)

Thus in 1.5 sec horizontal distance the rooster will move a distance of 1.5*15 = 22.5m

and will fall a distance of h' = 5(1.5) + 10(1.5)²/2          (s = ut + at²/2)

h' = 18.75m

Thus the rooster will regain its flight at a height of 73.42 - 18.75 = 54.67m above the ground and at a distance of 355.8 - 22.5 = 333.3m away from the barn.