Question

Pelicans tuck their wings and free-fall straight down whei} diving for fish. Suppose a pelican starts its dive from q height of 16.0m and cannot change its path once committed. If it takes a fish 0.20 s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fish is at the surface of the water.

Answer 1

Answer:

3.3 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16+0^2}\\\Rightarrow v=17.72\ m/s$

$v=u+at\\\Rightarrow 17.72=0+9.81\times t\\\Rightarrow \frac{17.72}{9.81}=t\\\Rightarrow t=1.81 \s$

The fish needs to see the pelican before 0.2 seconds in order to escape so the time pelican has is 1.81-0.2=1.61 seconds

In that time the pelican would have traveled

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 1.61+\frac{1}{2}\times 9.81\times 1.61^2\\\Rightarrow s=12.71\ m$

So, the pelican would be 16-12.71 = 3.3 m above the water.