An owl is carrying a mouse to the chicks in its nest. Its position at that

time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle30.0ºbelow the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

• Position of the mouse initially = 4.00m west and 12 m above the center of the nest.
• Diameter of the nest, d= 30.0 cm = 0.30 m
• The initial velocity of the owl flying towards the east, v_i= 3.50 m/s
• Angle below the horizontal when it drops the mouse, θ = 30.0°

We must determine the time the mouse requires to reach the nest’s level. So we have to resolve the initial velocity, v_i into its components.

We have the horizontal component of v­_i along the positive x-direction towards the east, from the figure:

$$\begin{gathered} \sin \theta =\frac{-V_{iy}}{V_i} \\ {V}_{iy}=-V_i\sin \theta \\ {V}_{iy}=-\sin \left({30}^{°}\right)\times 3.5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ {V}_{iy}=-1.75\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Thus v_iy, is negative because the direction is in the negative y-direction.

Also, we have ∆y = -12 m, a change in displacement in the negative y-direction and the acceleration, a­_­y = 9.8m/s

To find the time of flight, we have the equation:

$∆y=V_{iy}t+\frac{1}{2}{a}_y{t}^2$

On substituting the given values, we get:

$$\begin{gathered} -12=\left(1.75\right)t+\frac{1}{2}\left(9.8\right)t^2 \\ -12=\left(1.75\right)t+4.9t^2 \\ 4.9t^2+1.75t+12=0 \end{gathered}$$

Finding the root of the quadratic equation as:

$\begin{array}{rcl}t& =& \frac{-1.75\pm \sqrt{{\left(1.75\right)}^2-4\times 4.9\times \left(-12\right)}}{2\times 4.9}\\ & =& \frac{-1.75\pm \sqrt{3.06+235.2}}{9.8}\\ & =& \frac{-1.75\pm 15.43}{9.8}\\ & =& 1.40s\end{array}$

We have taken only the positive value because time is always positive.

This is the time required by the mouse to reach the horizontal level containing the nest.

Now we have to find the horizontal distance, so we need the horizontal component of the initial velocity, v_i. That is from the figure,

$\begin{array}{rcl}\cos \theta & =& \frac{V_{ix}}{V_i}\\ \cos \theta & =& \frac{V_{ix}}{3.5}\\ V_{ix}& =& 3.5\cos {30}^{°}\\ & =& 3.03\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Since the acceleration along the x-direction, a_y = 0 m/s, the velocity along the x-direction is also a constant.

$V_{ix}=V_{fx}=3.03\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Where v_fx,is the final velocity in the horizontal x-direction.

Therefore the average velocity in the horizontal direction:

$$\begin{gathered} V=\frac{x}{t} \\ V=\frac{V_{ix}+V_{fx}}{2} \\ V=3.03\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

On substituting the values,

$\begin{array}{rcl}x& =& 3.03\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\times 1.40s\\ & =& 4.24m\end{array}$

So $x=4.24m$ is the actual horizontal distance traveled by the mouse.

From the question, the radius of the nest is:

$\begin{array}{rcl}r& =& \frac{0.30}{2}\\ & =& 0.15m\end{array}$

Therefore the distance to the first end of the nest d₁ = 4 – 0.15 = 3.85 m.

Distance to the second end of the nest d₂ = 4 + 0.15 = 4.15 m.

The actual horizontal distance traveled by the mouse is .

So we can say that the mouse will not fall into the nest but outside the nest.