Final answer:
Calculating the horizontal velocity of the owl and the time it takes for the mouse to fall, the horizontal position of the mouse when it hits the ground is approximately 0.72 m east of the nest center. Since this distance is larger than half the nest's diameter, the owl is not lucky enough; the mouse will miss the nest.
Step-by-step explanation:
The question involves calculating the horizontal position of the mouse when it has fallen 12.0 m to determine if it hits the nest. As the owl is flying east at a velocity of 3.50 m/s, we will calculate the horizontal component of this velocity. To do so, we use the cosine component because the owl is flying at a 30.0° angle below the horizontal, resulting in a horizontal velocity (vx) of:
vx = vtotal × cos(θ)
vx = 3.50 m/s × cos(30.0°)
vx ≈ 3.03 m/s
Now, we use the kinematic equation for free fall to determine the time (t) it takes for the mouse to fall 12.0 m.
s = ½gt² → t = √(2s/g)
t = √(2 × 12.0 m / 9.81 m/s²)
t ≈ 1.56 s
Finally, we can find the displacement in the horizontal direction:
Δx = vx × tΔx = 3.03 m/s × 1.56 sΔx ≈ 4.72 m east
The mouse was initially 4.00 m west of the nest, so we add the displacement eastward. Hence, the horizontal position of the mouse when it has fallen 12.0 m:
horizontal position = initial position + displacementhorizontal position = -4.00 m + 4.72 mhorizontal position ≈ 0.72 m east
Since the nest diameter is 30.0 cm (0.3 m) and the horizontal position of the mouse is 0.72 m east, the mouse will miss the nest.