Answer:
s = 153.34 m
Explanation:
given,
speed of hawk flying = 11 m/s
altitude = 132 m
equation of parabolic trajectory
[tex]y = 132 -\dfrac{x^2}{33}[/tex]............at y = 0 x =66
[tex]ds=\sqrt{dx^2+dy^2}[/tex]
[tex]ds=dx\sqrt{1+(\dfrac{dy}{dx})^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{-2x}{33}[/tex]
[tex]ds=dx\sqrt{1+(\dfrac{-2x}{33} )^2}[/tex]
[tex]ds=dx\sqrt{1+(\dfrac{4x^2}{1089} )}[/tex]
integrating
[tex]s = \int _0^66\sqrt{1+(\dfrac{4x^2}{1089} )}[/tex]
using formula
[tex]\sqrt{x^2+a^2}=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}log|x+\sqrt{x^2+a^2}|[/tex]
s = 153.34 m
