## 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

Question

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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1 year 2021-09-02T07:49:49+00:00 1 Answers 4 views 0

1. a) 37.9 m/s

c) 7.56 s

Explanation:

a-c)

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

$$v_x=8.3 m/s$$

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

$$d_x = v_x t =(8.3)(2)=16.6 m$$

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

$$g=9.8 m/s^2$$ 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:

$$s=u_y t – \frac{1}{2}gt^2$$

where $$u_y = 0$$ is the initial vertical velocity. Solving for t,

$$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(280)}{9.8}}=7.56 s$$

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

$$v_y=\frac{s}{t}=\frac{280}{7.56}=37.0 m/s$$

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

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

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

$$v_x=8.3 m/s\\v_y=37.0 m/s$$

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

$$\theta=tan^{-1}(\frac{v_y}{v_x})$$

And substituting these values, we find:

$$\theta=tan^{-1}(\frac{37.0}{8.3})=77.4^{\circ}$$

And this angle is below the horizontal direction.

$$\theta=77.4\cdot \frac{2\pi}{360}=1.35 rad$$