Question

A hawk flying at 11 m/s at an altitude of 132 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation y = 132 − x2 33 until it hits the ground, where y is its height above the ground and x is its horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.

Answer 1

Answer:

s = 153.34 m

Explanation:

given,

speed of hawk flying = 11 m/s

altitude = 132 m

equation of parabolic trajectory

$y = 132 -\dfrac{x^2}{33}$............at y = 0    x =66

$ds=\sqrt{dx^2+dy^2}$

$ds=dx\sqrt{1+(\dfrac{dy}{dx})^2}$

$\dfrac{dy}{dx}=\dfrac{-2x}{33}$

$ds=dx\sqrt{1+(\dfrac{-2x}{33} )^2}$

$ds=dx\sqrt{1+(\dfrac{4x^2}{1089} )}$

integrating

$s = \int _0^66\sqrt{1+(\dfrac{4x^2}{1089} )}$

using formula

$\sqrt{x^2+a^2}=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}log|x+\sqrt{x^2+a^2}|$

s = 153.34 m