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## Homework Statement

A hawk flying at 2 m/s at an altitude of 80 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation below 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.

y = 80 - x^2/80

ans : ____m

## The Attempt at a Solution

arrggggg: here it is :

y' = -1/40x

1 + (f '(x)) = 1 + 1/1600 x^2

[tex]\int\sqrt{1 + 1/1600*x^2}[/tex]

=

1/40[tex]\int \sqrt{1600 + x^2}[/tex]

using trig sub and letting x = 40tan( [tex]\Theta[/tex] )

=

40 [tex]\int sec( \Theta )^3[/tex]

=

[tex]\frac{40}{41}[/tex][tex] sec(\vartheta)tan(\vartheta) + ln| sec(\vartheta)tan(\vartheta) | [/tex]

so,

1/40[tex]\int \sqrt{1600 + x^2}[/tex] = [tex]\frac{40}{41}[/tex][tex] sec(\vartheta)tan(\vartheta) + ln| sec(\vartheta)tan(\vartheta) | [/tex]

I am very lost if this is the integral. Any help please....