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?

Answer 1

a) 37.9 m/s

b) 1.35 rad below horizontal

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.

Converting into radians,

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