An Osprey can fly horizontally (not diving) at a maximum speed of 70.0 km/hr. The bird takes off from rest at the edge of a cliff, heading east, and accelerates at a rate of 1.21 m/s2. The osprey can decelerate at higher rate of 2.43 m/s2. The Osprey can reach top speeds of about 84 mph. When doing so, it does not flap its wings. Rather, it relies on its gravitational force to accelerate it downwards. a. Assuming that it starts from rest, and assuming that drag forces are slim, how long will it take for the osprey to reach its top speed? b. How far does it travel during this time? c. Near the top speed, drag forces become important, and in fact the terminal velocity is determined by the drag force. Assuming that the osprey's body can be modeled as a cylinder, with its cross-sectional area equal to that of a circle with radius of 10 cm, what is the coefficient of drag C for a hawk?

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An Osprey can fly horizontally (not diving) at a maximum speed of 70.0 km/hr. The bird takes off from rest at the edge of a cliff, heading east, and accelerates at a rate of 1.21 m/s2. The osprey can decelerate at higher rate of 2.43 m/s2. The Osprey can reach top speeds of about 84 mph. When doing so, it does not flap its wings. Rather, it relies on its gravitational force to accelerate it downwards.

a. Assuming that it starts from rest, and assuming that drag forces are slim, how long will it take for the osprey to reach its top speed?

b. How far does it travel during this time?

c. Near the top speed, drag forces become important, and in fact the terminal velocity is determined by the drag force. Assuming that the osprey's body can be modeled as a cylinder, with its cross-sectional area equal to that of a circle with radius of 10 cm, what is the coefficient of drag C for a hawk?

Expert Solution
Step 1

Converting the velocities  in m/s of =  70 km/h= 7000/3600=1.944m/s

acceleration=a= 1.21m/s2    and decelerate =d=2.43m/s2 

 its top speed is 84mph = 84/3600 = .0233 m/s  

  (b)   v= u +1.21xt

          t= .0192m/s     (assuming that it  is accelerating  downward)

On deceleration:

          .0233=0+9.8t

t=.0023 s      (as it reaches top  it can accelerate or decelerate)      

  (c) It travels a distance s= ut+1/2 (9.8x.0023x.0023) =.00026m.

 

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