Question

the minimum flying speee for a bird called a house martin is 9.0ms^-1. it reaches this speed by falling from its nest before swooping away. calculate the minimum distance its nest must be away above the ground.​

Answer 1

The nest must be about 4.15 meters above ground

Step-by-step explanation:

Use the velocity equation under accelerated motion (acceleration of gravity ):

`v_f=v_i+a\,*\,t`

which for this case has initial velocity = 0 (falls from the nest), final velocity = 9 m/s, and a = 9.8 m/s^2, then we can find the time needed in air while falling to reach the required speed:

`v_f=v_i+a\,*\,t\\9=0+9.8\,t\\t=(9)/(9.8) \, sec\\t \approx 0.92\,\, sec`

We now use this time value to find the distance covered in free fall during 0.92 seconds:

`d=(1)/(2) \,9.8 \,t^2=4.9\,(0.92)^2=4.147 \,meters\approx 4.15\,meters`