Part A: A bird glides horizontally (no wing flap) at 3m/sec elevation. It then g

Question

Part A: A bird glides horizontally (no wing flap) at 3m/sec elevation. It then glides downward for a ways, and then resumes horizontal gliding at a level of 2m below its original gliding level. If it never flaps its wings, and we ignore any retarding air friction in the face of the bird, how fast should it be gliding at the end of all of this? Show calculation or reasoning.

Part B: Same as above, except that the bird uses its flapping wings as brakes during its descent. Now how fast should it be going at the end? Show calculation or reasoning. Thank you so much!!

Explanation / Answer

its horizontal velocity will remain constant at 3 m/s

when it drops 2 m, its vertical veloicty will increase.

initial vertical velocity=0

vertical acceleration=9.8 m/s^2

distance travelled=2 m

then using the formula:

final velocity^2-initial velocity^2=2*9.8*distance

==> final velocity=6.261 m/s

hence net veloicty=sqrt(horizontal veloicty^2+vertical veloicty^2)=sqrt(3^2+6.261^2)=6.9426 m/s

part b:

as the deceleration due to breaking is not provided, it can not be calculated what its vertical velocity will be.

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