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## An owl is carrying a vole in its talons, flying in a horizontal direction at 8.3 m/s while 282 m above the ground. The vole wiggles free, an

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An owl is carrying a vole in its talons, flying in a horizontal direction at 8.3 m/s while 282 m above the ground. The vole wiggles free, and it takes the owl 2 s to respond. When it does respond, it dive at a constant speed in a straight line, catching the mouse 2 m from the ground (a) What is the owl’s dive speed? (b) What is the owl’s dive angle below the horizontal?(n radians) (c) How long, in seconds, does the mouse fall?

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Physics
1 year
2021-09-02T07:49:49+00:00
2021-09-02T07:49:49+00:00 1 Answers
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## Answers ( )

a) 37.9 m/sb) 1.35 radbelow horizontalc) 7.56 sExplanation:a-c)

At the beginning, both the owl and the vole are travelling in a horizontal direction at a speed of

[tex]v_x=8.3 m/s[/tex]

After the vole wiggles free, the owl takes 2 seconds to react; the horizontal distance covered by the owl during this time is

[tex]d_x = v_x t =(8.3)(2)=16.6 m[/tex]

The vertical motion of the wiggle is a free fall motion, so it is a uniformly accelerated motion with constant acceleration

[tex]g=9.8 m/s^2[/tex] in the downward direction

The wiggle falls from a height of h’ = 282 m to a height of h = 2 m, so the vertical displacement is

s = h’ – h = 282 – 2 = 280 m

The time it takes the wiggle to cover this distance is given by the suvat equation:

[tex]s=u_y t – \frac{1}{2}gt^2[/tex]

where [tex]u_y = 0[/tex] is the initial vertical velocity. Solving for t,

[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(280)}{9.8}}=7.56 s[/tex]

The owl must cover the vertical distance of 280 m in this time interval, so its vertical speed must be:

[tex]v_y=\frac{s}{t}=\frac{280}{7.56}=37.0 m/s[/tex]

Therefore, the speed of the owl during the dive is the resultant of the velocities in the two directions:

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{8.3^2+37.0^2}=37.9 m/s[/tex]

b)

In part a-c, we calculated that the components of the velocity of the owl in the horizontal and vertical direction, and they are

[tex]v_x=8.3 m/s\\v_y=37.0 m/s[/tex]

This means that the angle of the owl’s dive, with respect to the original horizontal direction, is

[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex]

And substituting these values, we find:

[tex]\theta=tan^{-1}(\frac{37.0}{8.3})=77.4^{\circ}[/tex]

And this angle is below the horizontal direction.

Converting into radians,

[tex]\theta=77.4\cdot \frac{2\pi}{360}=1.35 rad[/tex]