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

Answer:

The nest must be about  4.15 meters above ground

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:

$v_f=v_i+a\,*\,t\\9=0+9.8\,t\\t=\frac{9}{9.8} \, sec\\t \approx 0.92\,\, sec$

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

$d=\frac{1}{2} \,9.8 \,t^2=4.9\,(0.92)^2=4.147 \,meters\approx 4.15\,meters$